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 sys
from itertools import permutations

sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
f_inf = float('inf')
mod = 10 ** 9 + 7


def resolve():
    k = int(input())
    RC = [list(map(int, input().split())) for _ in range(k)]

    res = [["."] * 8 for _ in range(8)]
    for pattern in permutations(list(range(8))):
        XY = set(tuple((idx, num) for idx, num in enumerate(pattern)))
        flg = True
        for r, c in RC:
            if (r, c) not in XY:
                flg = False
                break
        if not flg:
            continue

        for x, y in XY:
            for i in range(1, 8):
                if (x + i, y + i) in XY or (x + i, y - i) in XY or (x - i, y + i) in XY or (x - i, y - i) in XY:
                    flg = False
                    break
            else:
                continue
            break
        if flg:
            for x, y in XY:
                res[x][y] = "Q"
            break

    for i in res:
        print("".join(i))


if __name__ == '__main__':
    resolve()

