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...
#define _GLIBCXX_DEBUG
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ALL(v) v.begin(), v.end()
typedef long long ll;

#include <bits/stdc++.h>
using namespace std;

int main(){
  int k;
  cin>>k;
  
  int r,c;
  vector<int> A(8),B(8);
  rep(i,8) A[i]=i;
  rep(i,8) B[i]=-1;
  rep(i,k){
    cin>>r>>c;
    B[r]=c;
  }
  
  char C[8][8];
  rep(i,8){
    rep(j,8) C[i][j]='.';
  }
  
  set<int> p,q;
  do{
    bool b=true;
    rep(i,8){
      if(B[i]!=-1 && B[i]!=A[i]) b=false;
    }
    if(b==false) continue;
    rep(i,8) p.insert(i+A[i]);
    rep(i,8) q.insert(i-A[i]);
    if(p.size()!=8 || q.size()!=8){
      p.clear(); q.clear();
      continue;
    }
    rep(i,8) C[i][A[i]]='Q';
    break;
  }while(next_permutation(ALL(A)));
  
  rep(i,8){
    rep(j,8) cout<<C[i][j];
    cout<<endl;
  }
  
  return 0;
}
