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 -*-

N = 8

row = [-1 for i in range(N)]
col = [False for i in range(N)]
board = [[False for i in range(N)] for j in range(N)]
pos = [False for i in range(2*N-1)]
neg = [False for i in range(2*N-1)]

num = int(input())

for i in range(num):
  x, y = list(map(int, input().split()))
  board[x][y] = True
  
def priboard():
  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):
      char = 'Q' if row[i] == j else '.'
      print(char, end="")
    print()
      
def back_track(k):
  if k == N:
    priboard()
    return 0
  for i in range(N):
    if (not col[i]) and (not pos[k+i]) and (not neg[k-i+N-1]):
      row[k] = i
      col[i] = True
      pos[k+i] = True
      neg[k-i+N-1] = True
      
      back_track(k+1)
      row[k] = -1
      col[i] = False
      pos[k+i] = False
      neg[k-i+N-1] = False

back_track(0)


