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...
from itertools import permutations

k = int(input())
rc = [list(map(int, input().split())) for _ in range(k)]

field = [["."] * 8 for _ in range(8)]

# 列
column = [i for i in range(8)]
# 行
row = [i for i in range(8)]

for r, perm_column in rc:
    column.remove(perm_column)
    row.remove(r)

    field[r][perm_column] = "Q"

dir_r = [1, 1, -1, -1]
dir_c = [1, -1, -1, 1]


def queen_check(field, perm_column):
    for i, j in zip(row, perm_column):
        for dr, dc in zip(dir_r, dir_c):
            r, c = i, j
            while True:
                r += dr
                c += dc

                if r < 0 or r >= 8 or c < 0 or c >= 8:
                    break

                if field[r][c] == "Q":
                    return False

    return True


for perm_column in permutations(column):
    for i, j in zip(row, perm_column):
        field[i][j] = "Q"

    if queen_check(field, perm_column):
        for f in field:
            print("".join(f))
    for i, j in zip(row, perm_column):
        field[i][j] = "."

