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 <vector>
#include <algorithm>
using namespace std;
using ll = long long;

int main(){
  int n = 8;
  int k;
  cin >> k;

  vector<pair<int,int>> q(k);
  for(int i=0;i<k;i++){
    int a,b;
    cin >> a >> b;
    q[i] = make_pair(a,b);
  }

  vector<int> order(n);
  for(int i=0;i<n;i++) order[i] = i;

  do{
    bool flag = false;
    for(int i=0;i<k;i++){
      if(order[q[i].first] != q[i].second){
        flag = true;
        break;
      }
    }
    if(flag) continue;

    flag = true;
    for(int i=0;i<n-1;i++){
      for(int j=i+1;j<n;j++){
        if(abs(j-i) == abs(order[j]-order[i])){
          flag = false;
        }else if(abs(j-i)==1 && abs(order[j]-order[i])==1){
          flag = false;
        }
      }
    }
    if(flag){
      break;
    }
  }while(next_permutation(order.begin(),order.end()));

  vector<vector<char>> ans(n,vector<char>(n));
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      if(order[i] == j){
        ans[i][j] = 'Q';
      }else{
        ans[i][j] = '.';
      }
    }
  }

  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      cout << ans[i][j];
    }
    cout << endl;
  }
}
