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...
k = int(input())
Qs = [tuple(map(int, input().split())) for _ in range(k)]
from itertools import permutations
Lis = [(i, j) for i in range(8) for j in range(8)]

def dfs(idx, lis):
    if idx==8:
        return lis
    Xs = set([x for x, y in lis])
    Ys = set([y for x, y in lis])
    Dif = set([x-y for x, y in lis])
    xDif = set([(8-x)-y for x, y in lis])
    ans = False
    for i in range(8):
        if i in Xs: continue
        for j in range(8):
            if j in Ys: continue
            if i-j in Dif: continue
            if (8-i)-j in xDif: continue
            lis.append((i, j))
            ans = dfs(idx+1, lis)
            if ans:
                return ans
            lis.pop()
    else:
        return False

chess = [['.']*8 for _ in range(8)]
for x, y in dfs(k, Qs):
    chess[x][y] = 'Q'
for i in range(8):
    print(''.join(chess[i]))
    
    
        
