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 PYTHON3):

Solution:
import sys
import math

sys.setrecursionlimit(300000)
MAXNUM = math.inf
MINNUM = -1 * math.inf


def getInt():
    return int(sys.stdin.readline().rstrip())


def getInts():
    return map(int, sys.stdin.readline().rstrip().split(" "))


def getString():
    return sys.stdin.readline().rstrip()


def printOutput(ans):
    for line in ans:
        for ele in line:
            sys.stdout.write(ele)
        sys.stdout.write("\n")


def emptySquares(g):
    total = 0
    firstEmpty = None
    for i in range(len(g)):
        for j in range(len(g[0])):
            if g[i][j] != "#":
                total += 1
                firstEmpty = (i, j)
    return total, firstEmpty


toDelete = []


def dfs(grid, visited, toSave, x, y):
    if (x, y) not in visited:
        visited[(x, y)] = True
        if toSave:
            toSave -= 1
        else:
            toDelete.append((x, y))

        for i, j in zip((-1, 1, 0, 0), (0, 0, 1, -1)):
            if (
                -1 < x + i < len(grid)
                and -1 < y + j < len(grid[0])
                and (x + i, y + j) not in visited
                and grid[x + i][y + j] != "#"
            ):
                toSave = dfs(grid, visited, toSave, x + i, y + j)
    return toSave


def iterativedfs(grid, toSave, x, y):
    stack = [(x, y)]
    visited = {}
    visited[(x,y)] = True
    toDelete = []
    while stack:
        x, y = stack.pop()

        if toSave != 0:
            toSave -= 1
        else:
            toDelete.append((x, y))

        for i, j in zip((-1, 1, 0, 0), (0, 0, 1, -1)):
            if (
                -1 < x + i < len(grid)
                and -1 < y + j < len(grid[0])
                and (x + i, y + j) not in visited
                and grid[x + i][y + j] != "#"
            ):
                stack.append((x + i, y + j))
                visited[(x+i, y+j)] = True

    return toDelete


def solve(n, m, k, grid):
    if k == 0:
        return grid
    empty, firstEmpty = emptySquares(grid)
    x, y = firstEmpty
    toDelete = iterativedfs(grid, empty - k, x, y)
    for a, b in toDelete:
        grid[a][b] = "X"
    return grid


def readinput():
    n, m, k = getInts()
    grid = []
    for _ in range(n):
        grid.append([i for i in getString()])
    printOutput(solve(n, m, k, grid))


readinput()