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...
N = 8
FREE = 1
NOT_FREE = -1


row = [FREE] * N
col = [FREE] * N
dpos = [FREE] * (2 * N - 1)
dneg = [FREE] * (2 * N - 1)

board = [[False] * N for _ in range(N)]

k = int(input())

for _ in range(k):
    r, c = map(int, input().split())
    board[r][c] = True


def slove(i):
    if i == N:
        print_board()
        return

    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] = NOT_FREE
        dpos[i + j] = NOT_FREE
        dneg[i - j + N - 1] = NOT_FREE

        slove(i + 1)

        row[i] = FREE
        col[j] = FREE
        dpos[i + j] = FREE
        dneg[i - j + N - 1] = FREE
       

def print_board():
    for i in range(N):
        for j in range(N):
            if board[i][j] and row[i] != j:
                return
    
    for i in range(N):
        for j in range(N):
            if row[i] == j:
                mark = "Q"
            else:
                mark = "."
            print(mark, end="")
        print()

slove(0)

