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 itertools

def check_p(p):
    #(a, b)の左上から右下への斜め線は(k, k+(b-a))
    for i in range(8):
        for j in range(i+1, 8):
            if p[j] - j == p[i] - i:
                return 0
    
    #(a, b)の右上から左下への斜め線は(a, b)について和一定
    for i in range(8):
        for j in range(i+1, 8):
            if p[j] + j == p[i] + i:
                return 0

    return 1

def print_p(p):
    for i in range(8):
        q = p[i]
        for j in range(8):
            if j != q:
                print(".", end = "")
            else:
                print("Q", end = "")
    
        print()
    return 0

k = int(input())
put_Q = list()
for i in range(k):
    put_Q.append(list(map(int, input().split())))

board = [["n" for _ in range(8)] for _ in range(8)]

error = 0
for p in itertools.permutations(list(range(8))):
    # print(p)
    for i in range(k):
        if p[put_Q[i][0]] != put_Q[i][1]:
            error = 1

    if error == 0:
        if check_p(p) == 1:
            print_p(p)
            break
    
    error = 0
