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>
using namespace std;

int main()
{
    int N; cin>>N;
    const int K=8;
    vector<int> pos(K,-1);
    for(int i=0;i<N;i++){
        int r,c; cin>>r>>c;
        pos[r]=c;
    }
    
    vector<int> p(K);
    for(int i=0;i<K;i++) p[i]=i;

    do{
        bool ok=true;
        for(int i=0;i<K;i++) if(pos[i]!=-1 and p[i]!=pos[i]) ok=false;
        for(int i=0;i<K-1;i++) for(int j=i+1;j<K;j++){
            if(abs(i-j)==abs(p[i]-p[j])) ok=false;
        }
        if(ok) break;
    }while(next_permutation(p.begin(),p.end()));

    vector<string> ans(K,string(K,'.'));
    for(int i=0;i<K;i++) ans[i][p[i]]='Q',cout<<ans[i]<<endl;
    return 0;
}
