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>
using namespace std;

enum{
  FREE,
  NOT_FREE
};

int D[8][8], row[8], col[8], dpos[15], dneg[15];

void print(){
  for(int i = 0; i < 8; i++){
    for(int j = 0; j < 8; j++) cout << (D[i][j] ? 'Q' : '.');
    cout << endl;
  }
}

void putQ(int i){
  if(i == 8){
    print();
    return;
  }
  if(row[i] == FREE){
    for(int j = 0; j < 8; j++){
      if(col[j] == NOT_FREE || dpos[i+j] == NOT_FREE || dneg[i-j+7] == NOT_FREE) continue;
      D[i][j] = 1;
      row[i] = col[j] = dpos[i+j] = dneg[i-j+7] = NOT_FREE;
      putQ(i+1);
      D[i][j] = 0;
      row[i] = col[j] = dpos[i+j] = dneg[i-j+7] = FREE;
    }
  }
  else putQ(i+1);
}

int main(){
  int n, r, c;
  cin >> n;
  for(int i = 0; i < n; i++){
    cin >> r >> c;
    D[r][c] = 1;
    row[r] = col[c] = dpos[r+c] = dneg[r-c+7] = NOT_FREE;
  }
  putQ(0);
  return 0;
}

