The goal of 8 Queens Problem is to put eight queens on a chess-board such that none of them threatens any of others. A queen threatens the squares in the same row, in the same column, or on the same diagonals as shown in the following figure.

<image>


For a given chess board where $k$ queens are already placed, find the solution of the 8 queens problem.

Constraints

* There is exactly one solution

Input

In the first line, an integer $k$ is given. In the following $k$ lines, each square where a queen is already placed is given by two integers $r$ and $c$. $r$ and $c$ respectively denotes the row number and the column number. The row/column numbers start with 0.

Output

Print a $8 \times 8$ chess board by strings where a square with a queen is represented by 'Q' and an empty square is represented by '.'.

Example

Input

2
2 2
5 3


Output

......Q.
Q.......
..Q.....
.......Q
.....Q..
...Q....
.Q......
....Q...
import itertools
import math
n = int(input())

placed = []

for i in range(n):
    a = list(map(int,input().split()))
    placed.append(a)

l = list(range(8))
perm = tuple(itertools.permutations(l))

for v in perm:
    flag = True
    
    for j in range(len(placed)):
        if v[placed[j][0]] != placed[j][1]:
            flag = False
            break 
    if flag == False:
        continue
    for i in range(8):
        k = 1
        # print(i)
        while  i+k<=7:
            # print(v,i,v[i],k,len(v))
            if i+k<=7:
                # print(i+k)
                if v[i+k] == v[i]+k or v[i+k] == v[i]-k:
                    flag = False
                    break
                # if v[i+k] == v[i]-k:
                #     flag = False
                #     break
            k += 1
        if flag == False:
            break
            
    if flag == True:
        # print(v)
        for r in range(8):
            for c in range(8):
                if c == v[r]:
                    print("Q",end = "")
                else:
                    print(".",end = "")
            print("")
                
        break
  
    # print(v[4],v)
    

