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;

int qcol[8] = {-1, -1, -1, -1, -1, -1, -1, -1};
bool flag = false;

void dfs(int i){
  if(i == 8){
    flag = true;
    return;
  }
  if(qcol[i] > 0){
    dfs(i+1);
    return;
  }
  for(int j=0; j<8; j++){
    bool fl = true;
    for(int k=0; k<8; k++){
      if(k==i || qcol[k]<0){
        continue;
      }
      if(j==qcol[k] || i-k==qcol[k]-j || i-k==j-qcol[k]){
        fl = false;
        break;
      }
    }
    if(fl){
      qcol[i] = j;
      dfs(i+1);
      if(flag){
        return;
      }
      else{
        qcol[i] = -1;
      }
    }
  }
  return;
}

int main(){
  int n; cin >> n;
  for(int i=0; i<n; i++){
    int x, y; cin >> x >> y;
    qcol[x] = y;
  }
  dfs(0);
  for(int i=0; i<8; i++){
    for(int j=0; j<8; j++){
      if(qcol[i] == j){
        cout << 'Q';
      }
      else{
        cout << '.';
      }
    }
    cout << "" << endl;
  }
}
