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
int1 = lambda x: int(x) - 1
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(500000)

N = int(readline())
RC = [list(map(int, readline().split())) for _ in range(N)]

for per in permutations(range(8), 8):
    board = [['.'] * 8 for _ in range(8)]
    flag = True
    for r, c in RC:
        if per[r] != c:
            flag = False
            break
    if flag:
        flag_ = True
        pos = []
        for i, p in enumerate(per):
            if any([abs(i - q[0]) == abs(p - q[1]) for q in pos]):
                flag_ = False
                break
            board[i][p] = 'Q'
            pos.append((i, p))
        if flag_:
            for r in board:
                print(''.join(r))

