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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;

int main(){
    int k;cin>>k;
    vector<int> v(8);
    vector<int> u(8,-1);
    vector<int> ans(8);
    for(int i=0;i<8;i++) v[i]=i;
    for(int i=0;i<k;i++){
        int r,c;cin>>r>>c;
        u[r]=c;
    }
    do{
        int flag=1;
        for(int i=0;i<8;i++){
            if(u[i]!=-1){
                if(u[i]!=v[i]) flag=0;
            }
        }
        if(flag){
            for(int i=0;i<8;i++){
                for(int j=0;j<8;j++){
                    if(i==j) continue;
                    if(abs(i-j)==abs(v[i]-v[j])) flag=0;
                }
            }
        }
        if(flag) ans=v;
    }while(next_permutation(v.begin(),v.end()));
    for(int i=0;i<8;i++){
        for(int j=0;j<8;j++){
            if(ans[i]==j) cout<<'Q';
            else cout<<'.';
        }
        cout<<endl;
    }
}
