Maze dfs倒行

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.

Example

Input

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

Output

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

Input

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

Output

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

一开始用的广搜，不对，广搜没啥合理性，实在是找不出道理，但是深搜可以，深搜可以把所有点连成一棵树，按顺序删除一些结点不会影响连通性，所以用递归
dfs直到四个方位都遍历过了，再把这个结点删除。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <queue>
#include <cmath>
using namespace std;
int n,m,k,rc=,sx,sy;
int dir[][]={,,,,,-,-,};
char mp[][];
int vis[][];
void dfs(int x,int y)
{
    if(x<||y<||x>=n||y>=m)return;
    if(mp[x][y]=='#'||vis[x][y])return ;
    vis[x][y]=;
    for(int i=;i<;i++)
    {
        int tx=x+dir[i][];
        int ty=y+dir[i][];
        dfs(tx,ty);
    }
    if(k)mp[x][y]='X',k--;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie();
    cin>>n>>m>>k;
    for(int i=;i<n;i++)
    {
        for(int j=;j<m;j++)
        {
            cin>>mp[i][j];
            if(mp[i][j]=='.')sx=i,sy=j;
            else vis[i][j]=;
        }
    }

    dfs(sx,sy);
    for(int i=;i<n;i++)
    {
        for(int j=;j<m;j++)
        putchar(mp[i][j]);

        putchar('\n');
    }
}