You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). A nonnegative integer A_{i,j} is written for each square (i,j).

You choose some of the squares so that each row and column contains at most K chosen squares. Under this constraint, calculate the maximum value of the sum of the integers written on the chosen squares. Additionally, calculate a way to choose squares that acheives the maximum.

Constraints

* 1 \leq N \leq 50
* 1 \leq K \leq N
* 0 \leq A_{i,j} \leq 10^9
* All values in Input are integer.

Input

Input is given from Standard Input in the following format:


N K
A_{1,1} A_{1,2} \cdots A_{1,N}
A_{2,1} A_{2,2} \cdots A_{2,N}
\vdots
A_{N,1} A_{N,2} \cdots A_{N,N}


Output

On the first line, print the maximum value of the sum of the integers written on the chosen squares.

On the next N lines, print a way that achieves the maximum.

Precisely, output the strings t_1,t_2,\cdots,t_N, that satisfies t_{i,j}=`X` if you choose (i,j) and t_{i,j}=`.` otherwise.

You may print any way to choose squares that maximizes the sum.

Examples

Input

3 1
5 3 2
1 4 8
7 6 9


Output

19
X..
..X
.X.


Input

3 2
10 10 1
10 10 1
1 1 10


Output

50
XX.
XX.
..X
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
// const int MOD = 998244353;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
const int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;

template <typename T, typename U>
struct PrimalDual {
  using Pui = pair<U, int>;

  struct Edge {
    int dst, rev;
    T cap;
    U cost;
    Edge(int dst, T cap, U cost, int rev) : dst(dst), cap(cap), cost(cost), rev(rev) {}
  };

  vector<vector<Edge>> graph;

  PrimalDual(int n, const T TINF, const U UINF) : n(n), TINF(TINF), UINF(UINF), graph(n), prev_v(n, -1), prev_e(n, -1), potential(n, 0), dist(n) {}

  void add_edge(int src, int dst, T cap, U cost) {
    has_negative_edge |= cost < 0;
    graph[src].emplace_back(dst, cap, cost, graph[dst].size());
    graph[dst].emplace_back(src, 0, -cost, graph[src].size() - 1);
  }

  U minimum_cost_flow(int s, int t, T flow) {
    U res = 0;
    if (has_negative_edge) {
      bellman_ford(s);
      if (dist[t] == UINF) return UINF;
      res += calc(s, t, flow);
    }
    while (flow > 0) {
      dijkstra(s);
      if (dist[t] == UINF) return UINF;
      res += calc(s, t, flow);
    }
    return res;
  }

  U minimum_cost_flow(int s, int t) {
    U res = 0;
    bellman_ford(s);
    if (potential[t] >= 0 || dist[t] == UINF) return res;
    T tmp = TINF;
    res += calc(s, t, tmp);
    while (true) {
      dijkstra(s);
      if (potential[t] >= 0 || dist[t] == UINF) return res;
      res += calc(s, t, tmp);
    }
  }

  pair<T, U> min_cost_max_flow(int s, int t, T flow) {
    T mx = flow;
    U cost = 0;
    if (has_negative_edge) {
      bellman_ford(s);
      if (dist[t] == UINF) return {mx - flow, cost};
      cost += calc(s, t, flow);
    }
    while (flow > 0) {
      dijkstra(s);
      if (dist[t] == UINF) return {mx - flow, cost};
      cost += calc(s, t, flow);
    }
    return {mx - flow, cost};
  }

private:
  int n;
  const T TINF;
  const U UINF;
  bool has_negative_edge = false;
  vector<int> prev_v, prev_e;
  vector<U> potential, dist;
  priority_queue<Pui, vector<Pui>, greater<Pui>> que;

  void bellman_ford(int s) {
    fill(ALL(dist), UINF);
    dist[s] = 0;
    bool is_updated = true;
    REP(step, n) {
      is_updated = false;
      REP(i, n) {
        if (dist[i] == UINF) continue;
        REP(j, graph[i].size()) {
          Edge e = graph[i][j];
          if (e.cap > 0 && dist[e.dst] > dist[i] + e.cost) {
            dist[e.dst] = dist[i] + e.cost;
            prev_v[e.dst] = i;
            prev_e[e.dst] = j;
            is_updated = true;
          }
        }
      }
      if (!is_updated) break;
    }
    assert(!is_updated);
    REP(i, n) {
      if (dist[i] != UINF) potential[i] += dist[i];
    }
  }

  void dijkstra(int s) {
    fill(ALL(dist), UINF);
    dist[s] = 0;
    que.emplace(0, s);
    while (!que.empty()) {
      Pui pr = que.top(); que.pop();
      int ver = pr.second;
      if (dist[ver] < pr.first) continue;
      REP(i, graph[ver].size()) {
        Edge e = graph[ver][i];
        U nx = dist[ver] + e.cost + potential[ver] - potential[e.dst];
        if (e.cap > 0 && dist[e.dst] > nx) {
          dist[e.dst] = nx;
          prev_v[e.dst] = ver;
          prev_e[e.dst] = i;
          que.emplace(dist[e.dst], e.dst);
        }
      }
    }
    REP(i, n) {
      if (dist[i] != UINF) potential[i] += dist[i];
    }
  }

  U calc(int s, int t, T &flow) {
    T f = flow;
    for (int v = t; v != s; v = prev_v[v]) f = min(f, graph[prev_v[v]][prev_e[v]].cap);
    flow -= f;
    for (int v = t; v != s; v = prev_v[v]) {
      Edge &e = graph[prev_v[v]][prev_e[v]];
      e.cap -= f;
      graph[v][e.rev].cap += f;
    }
    return potential[t] * f;
  }
};

int main() {
  int n, k; cin >> n >> k;
  PrimalDual<int, ll> pd(n * 2 + 2, INF, LINF);
  const int S = n * 2, T = n * 2 + 1;
  REP(i, n) {
    pd.add_edge(S, i, k, 0);
    pd.add_edge(n + i, T, k, 0);
  }
  REP(i, n) REP(j, n) {
    int a; cin >> a;
    pd.add_edge(i, n + j, 1, -a);
  }
  cout << -pd.minimum_cost_flow(S, T) << '\n';
  REP(i, n) {
    vector<bool> ans(n);
    for (auto e : pd.graph[i]) {
      if (n <= e.dst && e.dst < n * 2) ans[e.dst - n] = e.cap == 0;
    }
    REP(j, n) cout << (ans[j] ? 'X' : '.');
    cout << '\n';
  }
  return 0;
}
