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.

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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...
# 8 Queens Problem
[N, FREE, NOT_FREE] = [8, -1, 1]
initial_Q = [FREE for i in range(N)]
row = [FREE for i in range(N)]
col = [FREE for i in range(N)]
dneg = [FREE for i in range(N*2 - 1)]
dpos = [FREE for i in range(N*2 - 1)]

k = int(input())
for i in range(k):
    [r, c] = list(map(int, input().split()))
    initial_Q[r] = NOT_FREE
    row[r] = c
    [col[c], dpos[r + c], dneg[r - c + N - 1]] = [NOT_FREE, NOT_FREE, NOT_FREE]

def printBoard():
    global N, row
    for i in range(N):
        row_data = ["." for i in range(N)]
        row_data[row[i]] = "Q"
        print("".join(row_data))
    
def putQueen(i):
    global N, FREE, NOT_FREE
    global initial_Q
    global row, col, dneg, dpos
    if i == N:
        printBoard()
        return 1
    
    if initial_Q[i] != FREE:
        if putQueen(i + 1) == 1:
            return 1
        else:
            return 0
    else:
        for j in range(N):
            if col[j] == NOT_FREE or dpos[i + j] == NOT_FREE or dneg[i - j + N - 1] == NOT_FREE:
                continue
            row[i] = j
            [col[j], dpos[i + j], dneg[i - j + N - 1]] = [NOT_FREE, NOT_FREE, NOT_FREE]
            if putQueen(i + 1) == 1:
                return 1
            row[i] = FREE
            [col[j], dpos[i + j], dneg[i - j + N - 1]] = [FREE, FREE, FREE]

    return 0

putQueen(0)
