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.util.*;
   import java.io.*;
   import java.lang.Math.*;


  
   public class maze_337a
   {
      public static class coord {
      
         int x;
         int y;
         int d;
      
         public coord(int x, int y, int d)
         {
            this.x = x;
            this.y = y;
            this.d = d;
         }
      
      }
   
      
      private static String[][] maze;
      private static int[][] visited;
      private static int mazeRows;
      private static int mazeCols;
      private static Stack coordinates;
      private static int walls;
   
      public static void main(String[] args) {
      
         Scanner infile = null;
         try {
            infile = new Scanner((System.in)); //replace with System.in
         }
            catch(Exception e){}
         mazeRows  = infile.nextInt();	//number of tables
         mazeCols = infile.nextInt();	//number of tables
         walls = infile.nextInt();	//number of tables
         coordinates = new Stack();
      
      	//fill in that maze
         maze = new String[mazeRows][mazeCols];
         visited = new int[mazeRows][mazeCols];
         for(int i = 0; i < mazeRows; i++)
         {
            String input = infile.next();
         
            char[] myChar;
            myChar = input.toCharArray();
         
            for(int j = 0; j < mazeCols; j++)
            {
               visited[i][j] = 0;
               //System.out.println(i + " " + j);
               //System.out.println(myChar[j] + "");
               maze[i][j] = "" + myChar[j];
            }
         }
         //implement DFS here
         execute();
         	
         //print end result maze
      
               
      
      }
   
      private static void solver(int x, int y, int depth) //need a location to start solving at
      {
         //System.out.println(x + "\t" + y + "\t" + depth);
         if(x <0 || y < 0 || x >= mazeRows || y >= mazeCols ) 
            return;
         else if(visited[x][y] == 1 || maze[x][y].equals("#") )
            return; //done or viisted region
         else
         {
            visited[x][y] = 1;
            coordinates.push(new coord(x,y,depth)); //push depth of stack
         //floodfile DFS search
            //if(visited[x + 1][y] == 1 || maze[x + 1][y] != "#") 
            solver(x + 1, y, depth++);			
            //if(visited[x -1][y] == 1 || maze[x - 1][y] != "#") 
            solver(x - 1, y, (depth+1));			
            //if(visited[x][y + 1] == 0 || maze[x][y - 1] != "#") 
            solver(x, y + 1, (depth+1));			
            //if(visited[x][y - 1] == 0 || maze[x][y + 1] != "#") 
            solver(x, y - 1, (depth+1));			
         }
      }
      
      private static coord getCoord(String[][] maze)
      {
         for(int i = 0; i < maze.length; i++)
         {
            for(int j = 0; j < maze[0].length; j++)
            {
               //System.out.println(maze[i][j].equals("."));
               if(maze[i][j].equals("."))
               {
                  return new coord(i,j,0);
                  //System.out.print("HERE");
               }
            }
         }
         return null;
      }
   
      private static void execute() {
      
        //begin executino at point in code where you do have a period
         coord src = getCoord(maze);   
         if(src != null)
            solver(src.x, src.y, 0);
      		
      //now push coordinates from the stack
      //and set current maze to char value 'X'
      
         for(int k = 0; k < walls; k++)
         {
         //safety: ensure stack is not empty
            if(!coordinates.isEmpty())
            {
               coord temp = (coord)(coordinates.pop());
               //System.out.println(temp.x + "\t" + temp.y);
               maze[temp.x][temp.y] = "X";
            }
         }
      
      //end - print results
         print(maze);
      
      }
   
      public static void print(String[][] maze)
      {
         for(int i = 0; i < maze.length; i++)
         {
            for(int j = 0; j < maze[0].length; j++)
               System.out.print(maze[i][j] + "");
            System.out.println();
         }
      }
         
   }