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>
#include<math.h>
#include<ctime>

using namespace std;
using ll = long long;

bool solve(vector<int>masu, vector<int>point){
  for(int i = 0; i < 8; i++){
    if(point[i] != -1){
      if(point[i] != masu[i]) return false;
    }
  }
  vector<int> used_a(15, 0), used_b(15, 0);
  for(int i = 0; i < 8; i++){
    if(used_a[masu[i] - i + 7]) return false;
    if(used_b[masu[i] + i]) return false;
    used_a[masu[i] - i + 7] = 1;
    used_b[masu[i] + i] = 1;
  }
  return true;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int ti = clock();
  // start-----------------------------------------------
  int n; cin >> n;
  vector<int> masu(8);
  vector<int> point(8, -1);
  for(int i = 0; i < 8; i++) masu[i] = i;
  for(int i = 0; i < n; i++){
    int a, b;
    cin >> a >> b;
    point[a] = b;
  }
  do{
    if(solve(masu, point)) point = masu;
  }while(next_permutation(masu.begin(), masu.end()));
  for(int i = 0; i < 8; i++){
    for(int j = 0; j < 8; j++){
      if(point[i] == j)cout << 'Q';
      else cout << '.';
    }
    cout << endl;
  }
  // end-----------------------------------------------
  // cerr << 1.0 * (clock() - ti) / CLOCKS_PER_SEC << endl;
}

