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 <bits/stdc++.h>
#include<algorithm>
#define int long long
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define INF 1000000000000
int MOD=1000000007;

signed main() {
int n;
cin>>n;
vector<int>x(n);
vector<int>y(n);
rep(i,n)cin>>y[i]>>x[i];
vector<int>a(0);
rep(i,8)a.push_back(i);
do{
  int h=0;
rep(i,n){
  if(a[y[i]]!=x[i]){h++;}
}
if(h)continue;
rep(i,8){
  for(int j=i+1;j<8;j++){
    if(j-i==abs(a[j]-a[i])){h++;}
  }
}
if(!h){
  rep(i,8){
    rep(j,8){
      if(a[i]==j)cout<<'Q';
      else cout<<'.';
      if(j==7)cout<<endl;
    }
  }
  exit(0);
}
}while(next_permutation(a.begin(),a.end()));

}

