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...
#!/usr/bin/python3
import itertools

def print_boad(boad):
    b = [['.' for i in range(8)] for j in range(8)]
    for q in boad:
        b[q[0]][q[1]] = 'Q'

    for i in range(8):
        print(''.join(b[i]))


def find_next(base, exist_r, exist_c, exist_lt, exist_rt, invalid):
    if len(base) == 8:
        print_boad(base)
        exit()
    all_set = set(range(8))
    rest_r = all_set - exist_r
    rest_c = all_set - exist_c
    for r, c in itertools.product(rest_r, rest_c):
        if r - c in exist_lt:
            continue
        if r - (7 - c) in exist_rt:
            continue

        find_next(
                base | {(r, c)},
                exist_r | {r},
                exist_c | {c},
                exist_lt | {r - c},
                exist_rt | {r - (7 - c)},
                invalid)
        invalid.add((r, c))

    return False



# main

k = int(input())
boad = set()
exist_r = set()
exist_c = set()
exist_lt = set()
exist_rt = set()
invalid = set()
for i in range(k):
    r, c = map(int, input().split())
    boad.add((r, c))
    exist_r.add(r)
    exist_c.add(c)
    exist_lt.add(r - c)
    exist_rt.add(r - (7 - c))


find_next(boad, exist_r, exist_c, exist_lt, exist_rt, invalid)