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...
row = [True] * 8
col = [True] * 8
dpos = [True] * 15
dneg = [True] * 15


board = [['.', '.', '.', '.', '.', '.', '.', '.'] for i in range(8)]

import sys

file_input = sys.stdin

k = int(file_input.readline())

for line in file_input:
    r, c = map(int, line.split())
    row[r] = "INIT"
    col[c] = "INIT"
    dpos[r + c] = False
    dneg[r + (7 - c)] = False
    board[r][c] = 'Q'

def dfs(i = 0):
    if i == 8:
        for line in board:
            print(*line, sep='')
        return
    if row[i] == "INIT":
        dfs(i + 1)
    elif row[i]:
        for j in range(8):
            if col[j] == "INIT":
                pass
            elif col[j] and dpos[i + j] and dneg[i + (7 - j)]:
                row[i] = False
                col[j] = False
                dpos[i + j] = False
                dneg[i + (7 - j)] = False
                board[i][j] = 'Q'
                dfs(i + 1)
                row[i] = True
                col[j] = True
                dpos[i + j] = True
                dneg[i + (7 - j)] = True
                board[i][j] = '.'

dfs()