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...
# -*- coding: utf-8 -*-


if __name__ == '__main__':

    N = 8
    FREE = -1
    NOT_FREE = 1
    row = [FREE] * N
    col = [FREE] * N
    dpos = [FREE] * (2 * N - 1)
    dneg = [FREE] * (2 * N - 1)
    X = [["." for _ in range(N)] for _ in range(N)]

    k = int(input())
    for _ in range(k):
        i, j = map(int, input().split())
        X[i][j] = "Q"
        row[i] = col[j] = dpos[i + j] = dneg[i - j + N - 1] = NOT_FREE

    def printBoard(X):
        for i in range(N):
            print("".join(X[i]))

    def recursive(i):
        if i == N:
            printBoard(X)
            return
        if "Q" in X[i]:
            recursive(i + 1)

        for j in range(N):
            if NOT_FREE in [row[i], col[j], dpos[i + j], dneg[i - j + N - 1]]:
                continue
            else:
                X[i][j] = "Q"
                row[i] = col[j] = dpos[i + j] = dneg[i - j + N - 1] = NOT_FREE
            recursive(i + 1)
            X[i][j] = "."
            row[i] = col[j] = dpos[i + j] = dneg[i - j + N - 1] = FREE

    recursive(0)

