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
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
# from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
# import numpy as np
starttime = time.time()
# import numpy as np
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]

try:
    # sys.setrecursionlimit(int(pow(10,5)))
    sys.stdin = open("input.txt", "r")
    # sys.stdout = open("../output.txt", "w")
except:
    pass
global n,m,k,s
n,m,k=L()
A=[list(input()) for i in range(n)]

def bfs(i,j):
    global n,m,k,s
    vis=[[0 for i in range(m)] for j in range(n)]
    q=[[i,j]]
    vis[i][j]=1
    cnt=1
    while(q):
        x,y=q[0]
        del(q[0])
        if vis[x][y]>s-k:
            A[x][y]="X"
        row=[-1,0,0,1]
        col=[0,-1,1,0]
        for i in range(4):
            a=x+row[i]
            b=y+col[i]
            if a>=0 and b>=0 and a<n and b<m and vis[a][b]==0 and A[a][b]==".":
                cnt+=1
                vis[a][b]=cnt
                q.append([a,b])
s=0
for i in range(n):
    s+=(A[i].count("."))

for i in range(n):
    for j in range(m):
        if A[i][j]==".":
            bfs(i,j)
            print("\n".join("".join(ele) for ele in A))
            exit()


endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")