Problem: Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.

Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.

Input

The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.

Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.

Output

Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").

It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.

Examples

Input

3 4 2
#..#
..#.
#...


Output

#.X#
X.#.
#...


Input

5 4 5
#...
#.#.
.#..
...#
.#.#


Output

#XXX
#X#.
X#..
...#
.#.#

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T GCD(T a, T b) {
  a = abs(a);
  b = abs(b);
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <typename T>
inline T LCM(T x, T y) {
  T tp = GCD(x, y);
  if ((x / tp) * 1. * y > 9e18) return 9e18;
  return (x / tp) * y;
}
template <typename T>
inline T BigMod(T A, T B, T M) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M) {
  return BigMod(A, M - 2, M);
}
long long gcdr(long long a, long long b) {
  if (a == 0) return b;
  return gcdr(b % a, a);
}
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void FastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
}
char arr[550][550];
long long n, m, k;
long long findd = 0;
long long dx[] = {0, 0, 1, -1};
long long dy[] = {1, -1, 0, 0};
long long visited[550][550] = {0};
long long node, comp = 0;
bool dfs(long long nodex, long long nodey) {
  long long add = 1;
  node++;
  long long ans = false;
  visited[nodex][nodey] = 1;
  if (node == findd) {
    comp = 1;
    return true;
  }
  long long i, j;
  for (i = 0; i < 4 && !comp; i++) {
    long long u = nodex + dx[i];
    long long v = nodey + dy[i];
    if (u < n && u >= 0 && v < m && v >= 0) {
      if (arr[u][v] == '.' && visited[u][v] == 0) {
        ans = ans | dfs(u, v);
      }
    }
  }
  return ans;
}
int main() {
  long long t, i, j, x, y;
  t = 1;
  while (t--) {
    scanf("%lld", &n);
    scanf("%lld %lld", &m, &k);
    for (i = 0; i < n; i++) {
      cin >> arr[i];
    }
    long long emptyy = 0;
    long long lasti, lastj;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (arr[i][j] == '.') {
          emptyy++;
          lasti = i;
          lastj = j;
        }
      }
    }
    findd = emptyy - k;
    dfs(lasti, lastj);
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (visited[i][j] == 0 && arr[i][j] == '.') {
          cout << 'X';
        } else
          cout << arr[i][j];
      }
      cout << endl;
    }
  }
  return 0;
}