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<cmath>
#include<algorithm>
#include<vector>

using namespace std;

int main(){

    int k;
    cin >>k;
    
    int r[k],c[k];
    for(int i=0;i<k;i++){
        cin >> r[i];
        cin >> c[i];
    }

    int o[8]={0,1,2,3,4,5,6,7};
    int a[8]={};

    do{
        int d=0;
        for(int i=0;i<k;i++){
            if(o[r[i]]==c[i])d++;
        }
        if(d!=k)continue;

        long long int e=1;
        for(int i=0;i<7;i++){
            for(int j=i+1;j<8;j++){
                e=e*(i-j-o[i]+o[j])*(i+o[i]-j-o[j]);
            }
        }

        if(e!=0){
            for(int i=0;i<8;i++)a[i]=o[i];
            break;
        }
    }while(next_permutation(o,o+8));

    vector<string> t(8);


    for(int i=0;i<8;i++){
        for(int j=0;j<8;j++){
            if(a[i]==j)t[i].push_back('Q');
            else t[i].push_back('.');
        }
    }

    for(int i=0;i<8;i++)cout << t[i] << endl;


}
