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.BufferedWriter;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;

public class Main
{

	private static BufferedReader in;
	private static BufferedWriter out;
	private static int n;
	private static int m;
	private static int k;
	private static boolean[][] free;
	private static boolean[][] explored;
	static int[] dx = new int[]
	{ 1, -1, 0, 0 };
	static int[] dy = new int[]
	{ 0, 0, 1, -1 };

	public static void main(String[] args) throws NumberFormatException, IOException
	{
		// streams
		boolean file = false;
		if (file)
			in = new BufferedReader(new FileReader("input.txt"));
		else
			in = new BufferedReader(new InputStreamReader(System.in));
		out = new BufferedWriter(new OutputStreamWriter(System.out));
		// out = new BufferedWriter(new FileWriter("output.txt"));
		StringTokenizer tok;

		// read params
		tok = new StringTokenizer(in.readLine());
		n = Integer.parseInt(tok.nextToken());
		m = Integer.parseInt(tok.nextToken());
		k = Integer.parseInt(tok.nextToken());
		free = new boolean[n][m];
		int freeCount = 0;

		// read grid
		for (int i = 0; i < n; i++)
		{
			String str = in.readLine();
			for (int j = 0; j < m; j++)
				if (str.charAt(j) == '.')
				{
					free[i][j] = true;
					freeCount++;
				}
		}

		// get a free node and bfs from it
		Node source = getFreeNode();

		// explore all free nodes - k
		int req = freeCount - k;
		explored = new boolean[n][m];
		bfs(source, req);

		// print output
		for (int i = 0; i < n; i++)
		{
			StringBuilder strb = new StringBuilder();
			for (int j = 0; j < m; j++)
				if (!free[i][j])
					strb.append("#");
				else if (free[i][j] && !explored[i][j])
					strb.append("X");
				else
					strb.append(".");
			System.out.println(strb.toString());
		}

		out.flush();
	}

	private static void bfs(Node source, int req)
	{
		// put the source
		Queue<Node> q = new LinkedList<Node>();
		q.add(source);
		req--;
		explored[source.r][source.c] = true;

		// keep exploring while required is not reached
		while (req != 0)
		{
			// explored all the neighbours of this node
			Node node = q.poll();
			for (int i = 0; i < 4; i++)
			{
				int newR = node.r + dx[i];
				int newC = node.c + dy[i];
				if (valid(newR, newC) && !explored[newR][newC] && req != 0)
				{
					req--;
					explored[newR][newC] = true;
					q.add(new Node(newR, newC));
				}
			}

		}

	}

	private static boolean valid(int r, int c)
	{
		return r >= 0 && r < n && c >= 0 && c < m && free[r][c];
	}

	private static Node getFreeNode()
	{
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				if (free[i][j])
					return new Node(i, j);
		return null;
	}

}

class Node
{
	int r;
	int c;

	public Node(int r, int c)
	{
		this.r = r;
		this.c = c;
	}
}