Long is a huge fan of CFC (Codeforces Fried Chicken). But the price of CFC is increasing, so he decides to breed the chicken on his own farm.

His farm is presented by a rectangle grid with r rows and c columns. Some of these cells contain rice, others are empty. k chickens are living on his farm. The number of chickens is not greater than the number of cells with rice on the farm.

Long wants to give his chicken playgrounds by assigning these farm cells to his chickens. He would like to satisfy the following requirements:

  * Each cell of the farm is assigned to exactly one chicken. 
  * Each chicken is assigned at least one cell. 
  * The set of cells assigned to every chicken forms a connected area. More precisely, if two cells (x, y) and (u, v) are assigned to the same chicken, this chicken is able to walk from (x, y) to (u, v) by passing only its cells and moving from each cell to another cell sharing a side. 



Long also wants to prevent his chickens from fighting for food. Hence he wants the difference between the maximum and the minimum number of cells with rice assigned to a chicken to be as small as possible. Please help him.

Input

Each test contains multiple test cases. The first line contains the number of test cases T (1 ≤ T ≤ 2 ⋅ 10^4). Description of the test cases follows.

The first line of each test case contains three integers r, c and k (1 ≤ r, c ≤ 100, 1 ≤ k ≤ 62), representing the size of Long's farm and the number of chickens Long has. 

Each of the next r lines contains c characters, each is either "." or "R", representing an empty cell or a cell with rice. It is guaranteed that the number of chickens is not greater than the number of cells with rice on the farm.

It is guaranteed that the sum of r ⋅ c over all test cases does not exceed 2 ⋅ 10^4.

Output

For each test case, print r lines with c characters on each line. Each character should be either a lowercase English character, an uppercase English character, or a digit. Two characters should be equal if and only if the two corresponding cells are assigned to the same chicken. Uppercase and lowercase characters are considered different, so "A" and "a" belong to two different chickens.

If there are multiple optimal answers, print any.

Example

Input


4
3 5 3
..R..
...R.
....R
6 4 6
R..R
R..R
RRRR
RRRR
R..R
R..R
5 5 4
RRR..
R.R..
RRR..
R..R.
R...R
2 31 62
RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR
RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR


Output


11122
22223
33333
aacc
aBBc
aBBc
CbbA
CbbA
CCAA
11114
22244
32444
33344
33334
abcdefghijklmnopqrstuvwxyzABCDE
FGHIJKLMNOPQRSTUVWXYZ0123456789

Note

These pictures explain the sample output. Each color represents one chicken. Cells filled with patterns (not solid colors) contain rice.

In the first test case, each chicken has one cell with rice. Hence, the difference between the maximum and the minimum number of cells with rice assigned to a chicken is 0.

<image>

In the second test case, there are 4 chickens with 3 cells of rice, and 2 chickens with 2 cells of rice. Hence, the difference between the maximum and the minimum number of cells with rice assigned to a chicken is 3 - 2 = 1.

<image>

In the third test case, each chicken has 3 cells with rice. <image>

In the last test case, since there are 62 chicken with exactly 62 cells of rice, each chicken must be assigned to exactly one cell. The sample output is one of the possible way.
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1LL << 60;
string operator*(const string& s, int k) {
  if (k == 0) return "";
  string p = (s + s) * (k / 2);
  if (k % 2 == 1) p += s;
  return p;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
struct Edge {
  long long to, cap, rev;
  Edge(long long _to, long long _cap, long long _rev) {
    to = _to;
    cap = _cap;
    rev = _rev;
  }
};
void add_edge(vector<vector<Edge>>& G, long long from, long long to,
              long long cap, bool revFlag, long long revCap) {
  G[from].push_back(Edge(to, cap, (long long)G[to].size()));
  if (revFlag)
    G[to].push_back(Edge(from, revCap, (long long)G[from].size() - 1));
}
long long max_flow_dfs(vector<vector<Edge>>& G, long long v, long long t,
                       long long f, vector<bool>& used) {
  if (v == t) return f;
  used[v] = true;
  for (int i = 0; i < G[v].size(); ++i) {
    Edge& e = G[v][i];
    if (!used[e.to] && e.cap > 0) {
      long long d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
      if (d > 0) {
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
long long max_flow(vector<vector<Edge>>& G, long long s, long long t) {
  long long flow = 0;
  for (;;) {
    vector<bool> used(G.size());
    for (int(i) = (int)(0); i < (int)(used.size()); i++) used[i] = false;
    long long f = max_flow_dfs(G, s, t, INF, used);
    if (f == 0) {
      return flow;
    }
    flow += f;
  }
}
void BellmanFord(vector<vector<Edge>>& G, long long s, vector<long long>& d,
                 vector<long long>& negative) {
  d.resize(G.size());
  negative.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i] = INF;
  for (int(i) = (int)(0); i < (int)(d.size()); i++) negative[i] = false;
  d[s] = 0;
  for (int(k) = (int)(0); k < (int)(G.size() - 1); k++) {
    for (int(i) = (int)(0); i < (int)(G.size()); i++) {
      for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
        }
      }
    }
  }
  for (int(k) = (int)(0); k < (int)(G.size() - 1); k++) {
    for (int(i) = (int)(0); i < (int)(G.size()); i++) {
      for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
          negative[G[i][j].to] = true;
        }
        if (negative[i] == true) negative[G[i][j].to] = true;
      }
    }
  }
}
void Dijkstra(vector<vector<Edge>>& G, long long s, vector<long long>& d) {
  d.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i] = INF;
  d[s] = 0;
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      q;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    pair<long long, long long> a = q.top();
    q.pop();
    if (d[a.second] < a.first) continue;
    for (int(i) = (int)(0); i < (int)(G[a.second].size()); i++) {
      Edge e = G[a.second][i];
      if (d[e.to] > d[a.second] + e.cap) {
        d[e.to] = d[a.second] + e.cap;
        q.push(make_pair(d[e.to], e.to));
      }
    }
  }
}
void WarshallFloyd(vector<vector<Edge>>& G, vector<vector<long long>>& d) {
  d.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i].resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) {
    for (int(j) = (int)(0); j < (int)(d[i].size()); j++) {
      d[i][j] = INF;
    }
  }
  for (int(i) = (int)(0); i < (int)(G.size()); i++) {
    for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
      d[i][G[i][j].to] = G[i][j].cap;
    }
  }
  for (int(i) = (int)(0); i < (int)(G.size()); i++) {
    for (int(j) = (int)(0); j < (int)(G.size()); j++) {
      for (int(k) = (int)(0); k < (int)(G.size()); k++) {
        chmin(d[j][k], d[j][i] + d[i][k]);
      }
    }
  }
}
bool tsort(vector<vector<Edge>>& graph, vector<int>& order) {
  int n = graph.size(), k = 0;
  vector<long long> in(n);
  for (auto& es : graph)
    for (auto& e : es) in[e.to]++;
  priority_queue<long long, vector<long long>, greater<long long>> que;
  for (int(i) = (int)(0); i < (int)(n); i++)
    if (in[i] == 0) que.push(i);
  while (que.size()) {
    int v = que.top();
    que.pop();
    order.push_back(v);
    for (auto& e : graph[v])
      if (--in[e.to] == 0) que.push(e.to);
  }
  if (order.size() != n)
    return false;
  else
    return true;
}
class lca {
 public:
  const int n = 0;
  const int log2_n = 0;
  std::vector<std::vector<int>> parent;
  std::vector<int> depth;
  lca() {}
  lca(const vector<vector<Edge>>& g, int root)
      : n(g.size()),
        log2_n(log2(n) + 1),
        parent(log2_n, std::vector<int>(n)),
        depth(n) {
    dfs(g, root, -1, 0);
    for (int k = 0; k + 1 < log2_n; k++) {
      for (int v = 0; v < (int)g.size(); v++) {
        if (parent[k][v] < 0)
          parent[k + 1][v] = -1;
        else
          parent[k + 1][v] = parent[k][parent[k][v]];
      }
    }
  }
  void dfs(const vector<vector<Edge>>& g, int v, int p, int d) {
    parent[0][v] = p;
    depth[v] = d;
    for (auto& e : g[v]) {
      if (e.to != p) dfs(g, e.to, v, d + 1);
    }
  }
  int get(int u, int v) {
    if (depth[u] > depth[v]) std::swap(u, v);
    for (int k = 0; k < log2_n; k++) {
      if ((depth[v] - depth[u]) >> k & 1) {
        v = parent[k][v];
      }
    }
    if (u == v) return u;
    for (int k = log2_n - 1; k >= 0; k--) {
      if (parent[k][u] != parent[k][v]) {
        u = parent[k][u];
        v = parent[k][v];
      }
    }
    return parent[0][u];
  }
};
void visit(const vector<vector<Edge>>& g, int v, vector<vector<long long>>& scc,
           stack<long long>& S, vector<long long>& inS, vector<long long>& low,
           vector<long long>& num, int& time) {
  low[v] = num[v] = ++time;
  S.push(v);
  inS[v] = true;
  for (decltype((g[v]).begin()) e = (g[v]).begin(); e != (g[v]).end(); ++e) {
    int w = e->to;
    if (num[w] == 0) {
      visit(g, w, scc, S, inS, low, num, time);
      low[v] = min(low[v], low[w]);
    } else if (inS[w])
      low[v] = min(low[v], num[w]);
  }
  if (low[v] == num[v]) {
    scc.push_back(vector<long long>());
    while (1) {
      int w = S.top();
      S.pop();
      inS[w] = false;
      scc.back().push_back(w);
      if (v == w) break;
    }
  }
}
void stronglyConnectedComponents(const vector<vector<Edge>>& g,
                                 vector<vector<long long>>& scc) {
  const int n = g.size();
  vector<long long> num(n), low(n);
  stack<long long> S;
  vector<long long> inS(n);
  int time = 0;
  for (int(u) = (int)(0); u < (int)(n); u++)
    if (num[u] == 0) visit(g, u, scc, S, inS, low, num, time);
}
class UnionFind {
  vector<int> data;
  long long num;

 public:
  UnionFind(int size) : data(size, -1), num(size) {}
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    num -= (x != y);
    return x != y;
  }
  bool findSet(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  int size(int x) { return -data[root(x)]; }
  int numSet() { return num; }
};
class SumSegTree {
 private:
  long long _sum(long long a, long long b, long long k, long long l,
                 long long r) {
    if (r <= a || b <= l) return 0;
    if (a <= l && r <= b)
      return dat[k];
    else {
      long long s1 = _sum(a, b, 2 * k + 1, l, (l + r) / 2);
      long long s2 = _sum(a, b, 2 * k + 2, (l + r) / 2, r);
      return s1 + s2;
    }
  }

 public:
  long long n, height;
  vector<long long> dat;
  SumSegTree(long long _n) {
    n = 1;
    height = 1;
    while (n < _n) {
      n *= 2;
      height++;
    }
    dat = vector<long long>(2 * n - 1, 0);
  }
  void add(long long i, long long x) {
    i += n - 1;
    dat[i] += x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] += x;
    }
  }
  long long sum(long long a, long long b) { return _sum(a, b, 0, 0, n); }
};
class RmqTree {
 private:
  long long _find(long long a, long long b, long long k, long long l,
                  long long r) {
    if (r <= a || b <= l) return INF;
    if (a <= l && r <= b)
      return dat[k];
    else {
      long long s1 = _find(a, b, 2 * k + 1, l, (l + r) / 2);
      long long s2 = _find(a, b, 2 * k + 2, (l + r) / 2, r);
      return min(s1, s2);
    }
  }

 public:
  long long n, height;
  vector<long long> dat;
  RmqTree(long long _n) {
    n = 1;
    height = 1;
    while (n < _n) {
      n *= 2;
      height++;
    }
    dat = vector<long long>(2 * n - 1, INF);
  }
  void update(long long i, long long x) {
    i += n - 1;
    dat[i] = x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
  }
  long long find(long long a, long long b) { return _find(a, b, 0, 0, n); }
};
void divisor(long long n, vector<long long>& ret) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(ret.begin(), ret.end());
}
vector<long long> lis_fast(const vector<long long>& a) {
  const long long n = a.size();
  vector<long long> A(n, INT_MAX);
  vector<long long> id(n);
  for (int i = 0; i < n; ++i) {
    id[i] = distance(A.begin(), lower_bound(A.begin(), A.end(), a[i]));
    A[id[i]] = a[i];
  }
  long long m = *max_element(id.begin(), id.end());
  vector<long long> b(m + 1);
  for (int i = n - 1; i >= 0; --i)
    if (id[i] == m) b[m--] = a[i];
  return b;
}
bool z_algorithm(string& str, vector<int>& z, long long s) {
  const int L = str.size();
  z.resize(str.size());
  for (int i = 1, left = 0, right = 0; i < L; i++) {
    if (i > right) {
      left = right = i;
      for (; right < L && str[right - left] == str[right]; right++)
        ;
      z[i] = right - left;
      right--;
    } else {
      int k = i - left;
      if (z[k] < right - i + 1) {
        z[i] = z[k];
      } else {
        left = i;
        for (; right < L && str[right - left] == str[right]; right++)
          ;
        z[i] = right - left;
        right--;
      }
    }
    if (z[i] == s) return true;
  }
  return false;
}
bool z_algorithm(string& str, vector<int>& z) {
  const int L = str.size();
  z.resize(str.size());
  for (int i = 1, left = 0, right = 0; i < L; i++) {
    if (i > right) {
      left = right = i;
      for (; right < L && str[right - left] == str[right]; right++)
        ;
      z[i] = right - left;
      right--;
    } else {
      int k = i - left;
      if (z[k] < right - i + 1) {
        z[i] = z[k];
      } else {
        left = i;
        for (; right < L && str[right - left] == str[right]; right++)
          ;
        z[i] = right - left;
        right--;
      }
    }
  }
  return true;
}
struct RollingHash {
  static const int base1 = 1007, base2 = 2009;
  static const int mod1 = 1000000007, mod2 = 1000000009;
  vector<long long> hash1, hash2, power1, power2;
  RollingHash(const string& S) {
    int n = (int)S.size();
    hash1.assign(n + 1, 0);
    hash2.assign(n + 1, 0);
    power1.assign(n + 1, 1);
    power2.assign(n + 1, 1);
    for (int i = 0; i < n; ++i) {
      hash1[i + 1] = (hash1[i] * base1 + S[i]) % mod1;
      hash2[i + 1] = (hash2[i] * base2 + S[i]) % mod2;
      power1[i + 1] = (power1[i] * base1) % mod1;
      power2[i + 1] = (power2[i] * base2) % mod2;
    }
  }
  inline pair<long long, long long> get(int l, int r) const {
    long long res1 = hash1[r] - hash1[l] * power1[r - l] % mod1;
    if (res1 < 0) res1 += mod1;
    long long res2 = hash2[r] - hash2[l] * power2[r - l] % mod2;
    if (res2 < 0) res2 += mod2;
    return {res1, res2};
  }
  inline int getLCP(int a, int b) const {
    int len = min((int)hash1.size() - a, (int)hash1.size() - b);
    int low = 0, high = len;
    while (high - low > 1) {
      int mid = (low + high) >> 1;
      if (get(a, a + mid) != get(b, b + mid))
        high = mid;
      else
        low = mid;
    }
    return low;
  }
};
template <typename T>
T mod_pow(T x, T n, T mod) {
  T res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long mod_inv(long long x, long long mod) {
  return mod_pow(x, mod - 2, mod);
}
class Combination {
 public:
  vector<long long> fact;
  vector<long long> inv;
  long long mod;
  long long mod_inv(long long x) {
    long long n = mod - 2LL;
    long long res = 1LL;
    while (n > 0) {
      if (n & 1) res = res * x % mod;
      x = x * x % mod;
      n >>= 1;
    }
    return res;
  }
  long long nCr(long long n, long long r) {
    return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;
  }
  long long nPr(long long n, long long r) {
    return (fact[n] * inv[n - r]) % mod;
  }
  long long nHr(long long n, long long r) { return nCr(r + n - 1, r); }
  Combination(long long n, long long _mod) {
    mod = _mod;
    fact.resize(n + 1);
    fact[0] = 1;
    for (int(i) = (int)(0); i < (int)(n); i++) {
      fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
    }
    inv.resize(n + 1);
    inv[n] = mod_inv(fact[n]);
    for (int i = n; i > 0; i--) {
      inv[i - 1] = inv[i] * i % mod;
    }
  }
};
long long gcd(long long m, long long n) {
  if (n == 0) return m;
  return gcd(n, m % n);
}
long long lcm(long long m, long long n) { return m / gcd(m, n) * n; }
vector<vector<long long>> mIdentity(long long n) {
  vector<vector<long long>> A(n, vector<long long>(n));
  for (int i = 0; i < n; ++i) A[i][i] = 1;
  return A;
}
vector<vector<long long>> mMul(const vector<vector<long long>>& A,
                               const vector<vector<long long>>& B) {
  vector<vector<long long>> C(A.size(), vector<long long>(B[0].size()));
  for (int i = 0; i < C.size(); ++i)
    for (int j = 0; j < C[i].size(); ++j)
      for (int k = 0; k < A[i].size(); ++k)
        (C[i][j] += (A[i][k] % 1000000007) * (B[k][j] % 1000000007)) %=
            1000000007;
  return C;
}
vector<vector<long long>> mPow(const vector<vector<long long>>& A,
                               long long e) {
  return e == 0       ? mIdentity(A.size())
         : e % 2 == 0 ? mPow(mMul(A, A), e / 2)
                      : mMul(A, mPow(A, e - 1));
}
template <class T>
class RectangleSum {
 public:
  vector<vector<T>> sum;
  T GetSum(int left, int right, int top, int bottom) {
    T res = sum[bottom][right];
    if (left > 0) res -= sum[bottom][left - 1];
    if (top > 0) res -= sum[top - 1][right];
    if (left > 0 && top > 0) res += sum[top - 1][left - 1];
    return res;
  }
  RectangleSum(const vector<vector<T>>& s, int h, int w) {
    sum.resize(h);
    for (int i = 0; i < h; i++) sum[i].resize(w, 0);
    for (int y = 0; y < h; y++) {
      for (int x = 0; x < w; x++) {
        sum[y][x] = s[y][x];
        if (y > 0) sum[y][x] += sum[y - 1][x];
        if (x > 0) sum[y][x] += sum[y][x - 1];
        if (y > 0 && x > 0) sum[y][x] -= sum[y - 1][x - 1];
      }
    }
  }
};
long long _garner(vector<long long>& xs, vector<long long>& mods) {
  int M = xs.size();
  vector<long long> coeffs(M, 1), constants(M, 0);
  for (int i = 0; i < M - 1; ++i) {
    long long mod_i = mods[i];
    long long v = (xs[i] - constants[i] + mod_i) % mod_i;
    v = (v * mod_pow(coeffs[i], mod_i - 2, mod_i)) % mod_i;
    for (int j = i + 1; j < M; j++) {
      long long mod_j = mods[j];
      constants[j] = (constants[j] + coeffs[j] * v) % mod_j;
      coeffs[j] = (coeffs[j] * mod_i) % mod_j;
    }
  }
  return constants.back();
}
template <typename T>
inline void bit_reverse(vector<T>& a) {
  int n = a.size();
  int i = 0;
  for (int j = 1; j < n - 1; ++j) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(a[i], a[j]);
  }
}
template <long long mod, long long primitive_root>
class NTT {
 public:
  long long get_mod() { return mod; }
  void _ntt(vector<long long>& a, int sign) {
    const int n = a.size();
    assert((n ^ (n & -n)) == 0);
    const long long g = primitive_root;
    long long tmp = (mod - 1) * mod_pow((long long)n, mod - 2, mod) % mod;
    long long h = mod_pow(g, tmp, mod);
    if (sign == -1) h = mod_pow(h, mod - 2, mod);
    bit_reverse(a);
    for (int m = 1; m < n; m <<= 1) {
      const int m2 = 2 * m;
      long long _base = mod_pow((long long)h, (long long)(n / m2), mod);
      long long _w = 1;
      for (int x = 0; x < m; ++x) {
        for (int s = x; s < n; s += m2) {
          long long u = a[s];
          long long d = (a[s + m] * _w) % mod;
          a[s] = (u + d) % mod;
          a[s + m] = (u - d + mod) % mod;
        }
        _w = (_w * _base) % mod;
      }
    }
  }
  void ntt(vector<long long>& input) { _ntt(input, 1); }
  void intt(vector<long long>& input) {
    _ntt(input, -1);
    const long long n_inv = mod_pow((long long)input.size(), mod - 2, mod);
    for (auto& x : input) x = (x * n_inv) % mod;
  }
  vector<long long> convolution(const vector<long long>& a,
                                const vector<long long>& b) {
    int result_size = a.size() + b.size() - 1;
    int n = 1;
    while (n < result_size) n <<= 1;
    vector<long long> _a = a, _b = b;
    _a.resize(n, 0);
    _b.resize(n, 0);
    ntt(_a);
    ntt(_b);
    for (int i = 0; i < n; ++i) _a[i] = (_a[i] * _b[i]) % mod;
    intt(_a);
    _a.resize(result_size);
    return _a;
  }
};
vector<long long> convolution_ntt(vector<long long>& a, vector<long long>& b,
                                  long long mod = 1224736769LL) {
  for (auto& x : a) x %= mod;
  for (auto& x : b) x %= mod;
  long long maxval = max(a.size(), b.size()) *
                     *max_element(a.begin(), a.end()) *
                     *max_element(b.begin(), b.end());
  if (maxval < 1224736769) {
    NTT<1224736769, 3> ntt3;
    return ntt3.convolution(a, b);
  }
  NTT<167772161, 3> ntt1;
  NTT<469762049, 3> ntt2;
  NTT<1224736769, 3> ntt3;
  vector<long long> x1 = ntt1.convolution(a, b);
  vector<long long> x2 = ntt2.convolution(a, b);
  vector<long long> x3 = ntt3.convolution(a, b);
  vector<long long> ret(x1.size());
  vector<long long> mods{167772161, 469762049, 1224736769, mod};
  for (int i = 0; i < x1.size(); ++i) {
    vector<long long> xs{x1[i], x2[i], x3[i], 0};
    ret[i] = _garner(xs, mods);
  }
  return ret;
}
int popcount3(int x) {
  x = (x & 0x55555555) + (x >> 1 & 0x55555555);
  x = (x & 0x33333333) + (x >> 2 & 0x33333333);
  x = (x & 0x0F0F0F0F) + (x >> 4 & 0x0F0F0F0F);
  x = (x & 0x00FF00FF) + (x >> 8 & 0x00FF00FF);
  x = (x & 0x0000FFFF) + (x >> 16 & 0x0000FFFF);
  return x;
}
template <typename T>
void rowReduction(vector<T> mat, vector<T>& basis) {
  for (auto e : mat) {
    for (auto b : basis) chmin(e, e ^ b);
    if (e) basis.push_back(e);
  }
  sort((basis).begin(), (basis).end(), greater<T>());
}
map<long long, char> o;
void solve() {
  long long r, c, k;
  cin >> r >> c >> k;
  long long rice = 0;
  vector<string> s(r);
  for (int(i) = (int)(0); i < (int)(r); i++) {
    cin >> s[i];
    for (int(j) = (int)(0); j < (int)(c); j++)
      if (s[i][j] == 'R') rice++;
  }
  long long base = rice / k;
  long long rest = rice % k;
  long long now = 0;
  long long target = base + (!!rest);
  rest -= (!!rest);
  vector<vector<char>> ans(r, vector<char>(c));
  for (int(i) = (int)(0); i < (int)(r); i++) {
    for (int(j) = (int)(0); j < (int)(c); j++) {
      long long co;
      if (i % 2 == 0)
        co = j;
      else
        co = c - j - 1;
      ans[i][co] = o[now];
      if (s[i][co] == 'R') {
        target--;
        if (target == 0) {
          if (now < k - 1) now++;
          target = base + (!!rest);
          rest -= (!!rest);
        }
      }
    }
  }
  for (int(i) = (int)(0); i < (int)(r); i++) {
    for (int(j) = (int)(0); j < (int)(c); j++) cout << ans[i][j];
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  for (int(i) = (int)(0); i < (int)(10); i++) o[i] = '0' + i;
  for (int(i) = (int)(0); i < (int)(26); i++) o[i + 10] = 'a' + i;
  for (int(i) = (int)(0); i < (int)(26); i++) o[i + 36] = 'A' + i;
  for (int(i) = (int)(0); i < (int)(t); i++) solve();
  return 0;
}
