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...
#include<iostream>
#include<algorithm>
using namespace std;

int n = 8;
int row[8] = {}, col[8] = {}, dpos[16] = {}, dneg[16] = {}; //0->free
int queen[8][8] = {}; //0->free

void printBoard(){
  
  for(int i = 0;i < n;i++){
    for(int j = 0;j < n;j++){
      if(queen[i][j] && row[i] != j) return ;
    }
  }
  
  for(int i = 0;i < n;i++){
    for(int j = 0;j < n;j++){
      if(row[i] == j) cout << 'Q';
      else cout << '.';
    }
    cout << endl;
  }
}

void putQueen(int i){
  
  if(i == n){
    printBoard();
    return ;
  }
  
  for(int j = 0;j < n;j++){
    if(col[j] || dpos[i + j] || dneg[i - j + n - 1]) continue;
    
    row[i] = j;
    col[j] = dpos[i + j] = dneg[i - j + n - 1] = 1; //not free
    
    putQueen(i + 1);
    
    col[j] = dpos[i + j] = dneg[i - j + n - 1] = 0; //free
  }
}

int main(){
  
  int k, r, c;
  
  cin >> k;
  for(int i = 0;i < k;i++){
    cin >> r >> c;
    queen[r][c] = 1;
  }
  
  putQueen(0);
  
  return 0;
}
