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 k;scanf("%d",&k);
    int y[k],x[k];
    for(int i=0;i<k;i++)scanf("%d%d",&y[i],&x[i]);
    vector<int> a(8);
    for(int i=0;i<8;i++)a[i]=i;
    do{
        vector<vector<int>> ans(8,vector<int>(8));
        bool check=1;
        for(int i=0;i<8;i++)ans[i][a[i]]=1;
        for(int i=0;i<k;i++)if(ans[y[i]][x[i]]==0)check=0;
        for(int i=0;i<8;i++)for(int j=i+1;j<8;j++){
            if(a[i]-i==a[j]-j or a[i]+i==a[j]+j)check=0;
        }
        if(check){
            for(int i=0;i<8;i++){
                for(int j=0;j<8;j++){
                    if(ans[i][j])printf("%s","Q");
                    else printf("%s",".");
                }
                puts("");
            }
        }
    }while(next_permutation(a.begin(),a.end()));
    return 0;
}
