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<utility>
#include<algorithm>
using namespace std;
#define ii pair<int,int>

int main(){
    int k;cin>>k;
    vector<ii> c(k);
    for(int i=0;i<k;i++){
        int x,y;cin>>x>>y;
        c[i]=make_pair(x,y);
    }
    vector<int> f(8);
    for(int i=0;i<8;i++) f[i]=i;
    do{
        bool flag=true;
        for(int i=0;i<k;i++){
            if(f[c[i].first]!=c[i].second){
                flag=false;
                break;
            }
        }
        for(int i=0;i<8;i++){
            for(int j=i+1;j<8;j++){
                if(abs(f[i]-f[j])==j-i){
                    flag=false;
                    break;
                }
            }
            if(!flag) break;
        }
        if(flag){
            for(int i=0;i<8;i++){
                for(int j=0;j<8;j++){
                    if(j!=f[i]) cout<<'.';
                    else cout<<'Q';
                }
                cout<<endl;
            }
            return 0;
        }
    }while(next_permutation(f.begin(),f.end()));
}
