You already know that Valery's favorite sport is biathlon. Due to your help, he learned to shoot without missing, and his skills are unmatched at the shooting range. But now a smaller task is to be performed, he should learn to complete the path fastest.

The track's map is represented by a rectangle n × m in size divided into squares. Each square is marked with a lowercase Latin letter (which means the type of the plot), with the exception of the starting square (it is marked with a capital Latin letters S) and the terminating square (it is marked with a capital Latin letter T). The time of movement from one square to another is equal to 1 minute. The time of movement within the cell can be neglected. We can move from the cell only to side-adjacent ones, but it is forbidden to go beyond the map edges. Also the following restriction is imposed on the path: it is not allowed to visit more than k different types of squares (squares of one type can be visited an infinite number of times). Squares marked with S and T have no type, so they are not counted. But S must be visited exactly once — at the very beginning, and T must be visited exactly once — at the very end.

Your task is to find the path from the square S to the square T that takes minimum time. Among all shortest paths you should choose the lexicographically minimal one. When comparing paths you should lexicographically represent them as a sequence of characters, that is, of plot types.

Input

The first input line contains three integers n, m and k (1 ≤ n, m ≤ 50, n·m ≥ 2, 1 ≤ k ≤ 4). Then n lines contain the map. Each line has the length of exactly m characters and consists of lowercase Latin letters and characters S and T. It is guaranteed that the map contains exactly one character S and exactly one character T.

Pretest 12 is one of the maximal tests for this problem.

Output

If there is a path that satisfies the condition, print it as a sequence of letters — the plot types. Otherwise, print "-1" (without quotes). You shouldn't print the character S in the beginning and T in the end.

Note that this sequence may be empty. This case is present in pretests. You can just print nothing or print one "End of line"-character. Both will be accepted.

Examples

Input

5 3 2
Sba
ccc
aac
ccc
abT


Output

bcccc


Input

3 4 1
Sxyy
yxxx
yyyT


Output

xxxx


Input

1 3 3
TyS


Output

y


Input

1 4 1
SxyT


Output

-1
#include <bits/stdc++.h>
using namespace std;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
struct kraw {
  int a;
  bool jest;
  int kier;
};
int main() {
  ios_base::sync_with_stdio(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<string> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<vector<kraw> > G(n * m);
  int start, meta;
  vector<char> lit(n * m);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (a[i][j] == 'S') start = i * m + j;
      if (a[i][j] == 'T') meta = i * m + j;
      lit[i * m + j] = a[i][j];
      for (int l = 0; l < 4; ++l) {
        int x = i + dx[l], y = j + dy[l];
        if (x >= 0 && x < n && y >= 0 && y < m) {
          G[i * m + j].push_back({x * m + y, true, 0});
          G[x * m + y].push_back({i * m + j, true, 0});
        }
      }
    }
  }
  string odp = "";
  for (int mask = 0; mask < (1 << 26); ++mask) {
    if (__builtin_popcount(mask) != k) continue;
    for (int i = 0; i < n * m; ++i) {
      for (int j = 0; j < G[i].size(); ++j) {
        G[i][j].jest = true;
        G[i][j].kier = 0;
      }
    }
    for (int i = 0; i < n * m; ++i)
      for (int j = 0; j < G[i].size(); ++j)
        if ((lit[i] != 'S' && lit[i] != 'T' &&
             (mask & (1 << (lit[i] - 'a'))) == 0) ||
            (lit[G[i][j].a] != 'S' && lit[G[i][j].a] != 'T' &&
             (mask & (1 << (lit[G[i][j].a] - 'a'))) == 0))
          G[i][j].jest = false;
    queue<int> Q;
    vector<int> odl(n * m, -1);
    Q.push(start);
    odl[start] = 0;
    while (!Q.empty()) {
      int akt = Q.front();
      Q.pop();
      for (int i = 0; i < G[akt].size(); ++i) {
        if (G[akt][i].jest && odl[G[akt][i].a] == -1) {
          Q.push(G[akt][i].a);
          odl[G[akt][i].a] = odl[akt] + 1;
        }
      }
    }
    if (odl[meta] == -1) continue;
    for (int i = 0; i < n * m; ++i) {
      for (int j = 0; j < G[i].size(); ++j) {
        if (G[i][j].jest) {
          if (odl[i] == odl[G[i][j].a]) G[i][j].jest = false;
          if (odl[i] < odl[G[i][j].a]) G[i][j].kier = 1;
          if (odl[i] > odl[G[i][j].a]) G[i][j].kier = -1;
        }
      }
    }
    vector<int> st_wy(n * m, 0);
    for (int i = 0; i < n * m; ++i)
      for (int j = 0; j < G[i].size(); ++j)
        if (G[i][j].jest && G[i][j].kier == 1) ++st_wy[i];
    for (int i = 0; i < n * m; ++i)
      if (st_wy[i] == 0) Q.push(i);
    while (!Q.empty()) {
      int akt = Q.front();
      Q.pop();
      if (akt == meta) continue;
      for (int i = 0; i < G[akt].size(); ++i) {
        if (G[akt][i].jest && G[akt][i].kier == -1) {
          int pop = G[akt][i].a;
          for (int j = 0; j < G[pop].size(); ++j)
            if (G[pop][j].a == akt && G[pop][j].jest) G[pop][j].jest = false;
          --st_wy[pop];
          if (st_wy[pop] == 0) Q.push(pop);
        }
      }
    }
    string path = "S";
    vector<int> a;
    a.push_back(start);
    while (path.size() < odl[meta]) {
      char l = 'z';
      for (int i = 0; i < a.size(); ++i)
        for (int j = 0; j < G[a[i]].size(); ++j)
          if (G[a[i]][j].jest && G[a[i]][j].kier == 1)
            l = min(l, lit[G[a[i]][j].a]);
      path += l;
      vector<int> b;
      set<int> S;
      for (int i = 0; i < a.size(); ++i) {
        for (int j = 0; j < G[a[i]].size(); ++j) {
          if (G[a[i]][j].jest && G[a[i]][j].kier == 1 &&
              lit[G[a[i]][j].a] == l && S.find(G[a[i]][j].a) == S.end()) {
            b.push_back(G[a[i]][j].a);
            S.insert(G[a[i]][j].a);
          }
        }
      }
      a = b;
    }
    path += "T";
    if (odp == "" || (odp != "" && path.size() < odp.size()) ||
        (odp != "" && path.size() == odp.size() && path < odp))
      odp = path;
  }
  if (odp == "")
    cout << "-1";
  else
    for (int i = 1; i < odp.size() - 1; ++i) cout << odp[i];
  return 0;
}
