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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

	static int n ,m,k;
	static char[][] a;
	static boolean[][] vis;
	static int[] di = {-1,0,1,0};
	static int[] dj = {0,1,0,-1};
	
	static void dfs(int i,int j)
	{
		vis[i][j] = true;
		if(k<=0)return;
		for (int k = 0; k < di.length; k++) 
		{
			int ni = i + di[k] , nj = j + dj[k];
			if(isValid(ni, nj) && !vis[ni][nj])
				dfs(ni,nj);
		}
		if(k-->0)a[i][j] = 'X';
		
	}
	
	static boolean isValid(int i,int j){return i>=0 && i<n && j>=0 &&j<m && a[i][j]!='#';}
	
	
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt(); m = sc.nextInt(); k = sc.nextInt();
		a = new char[n][m];
		vis = new boolean[n][m];
		for (int i = 0; i < n; i++) 
			a[i] = sc.nextLine().toCharArray();
		
		lop: for (int i = 0; i < n; i++) 
			for (int j = 0; j < m; j++) 
				if(a[i][j]=='.'){dfs(i,j);break lop;}
		
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < n; i++) 
		{
			for (int j = 0; j < m; j++) 
				sb.append(a[i][j]);
			sb.append("\n");
		}
		System.out.print(sb);
	}
	


	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;
		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}
		public Scanner(String s) throws FileNotFoundException {br = new BufferedReader(new FileReader(s));}
		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public String nextLine() throws IOException {return br.readLine();}		
		public long nextLong() throws IOException {return Long.parseLong(next());}		
		public double nextDouble() throws IOException {return Double.parseDouble(next());}	
		public char nextChar() throws IOException{return next().charAt(0);}
		public boolean ready() throws IOException {return br.ready();}
		public int[] nextIntArr() throws IOException{
			st = new StringTokenizer(br.readLine());
			int[] res = new int[st.countTokens()];
			for (int i = 0; i < res.length; i++)
				res[i] = nextInt();
			return res;
		}
		public char[] nextCharArr() throws IOException{
			st = new StringTokenizer(br.readLine());
			char[] res = new char[st.countTokens()];
			for (int i = 0; i < res.length; i++) 
				res[i] = nextChar();
			return res;
		}
	}

}