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 sumdiag(board, k):
    if (k >= 0):
        s = 0
        for i in range(len(board) - k):
            s += board[i][i + k]
        return s
    else:
        k = -k
        s = 0
        for i in range(len(board) - k):
            s += board[i + k][i]
        return s
           
def fliplr(board):
    f_board = []
    for i in range(len(board)):
        f_board.append(board[i][::-1])
    return f_board

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

for queen in itertools.permutations(range(8)):
    flag = 1
    for q in qs:
        if (queen[q[0]] != q[1]):
            flag = 0
    if (flag == 0):
        continue
    board = [[0 for i in range(8)] for j in range(8)]
    for i, qu in enumerate(queen):
        board[i][qu] = 1

    for k in range(-7,8):
        if (sumdiag(board, k=k) > 1):
            flag = 0
    f_board = fliplr(board)
    for k in range(-7, 8):
        if (sumdiag(f_board, k=k) > 1):
            flag = 0
    if (flag == 1):
        for i in range(8):
            for j in range(8):
                if (board[i][j] == 0):
                    print('.', end='')
                else:
                    print('Q', end='')
            print()
        break
