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 judge(board):
  ret=True
  pos=[]
  for i in range(8):
    for j in range(8):
      if board[i][j]=='Q':
        pos.append((i,j))
  for i in range(8):
    tx1,ty1=pos[i]
    for j in range(8):
      tx2,ty2=pos[j]
      if i==j:
        continue
      if ty1==ty2 or tx1==tx2 or abs(ty1-ty2)==abs(tx1-tx2):
        ret=False
  return ret

k=int(input())
board=[['.']*8 for _ in range(8)]
pos1=[0,1,2,3,4,5,6,7]
pos2=[0,1,2,3,4,5,6,7]
for _ in range(k):
  a,b=map(int,input().split())
  board[a][b]='Q'
  pos1.remove(b)
  pos2.remove(a)
for tpos in itertools.permutations(pos1):
  for i in range(len(tpos)):
    board[pos2[i]][tpos[i]]='Q'
  if judge(board)==True:
    break
  for i in range(len(tpos)):
    board[pos2[i]][tpos[i]]='.'
for i in range(8):
  print(''.join(board[i]))
