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<algorithm>
#include<cstring>
using namespace std;
int n;
int h[20],zx[20],yx[20],l[20];
int dt[10][10];
int f=0;
void dfs(int p){
    if(f) return;
    if(p==8){
        for(int i=0;i<8;i++){
            for(int j=0;j<8;j++){
                if(dt[i][j]) cout<<"Q";
                else cout<<'.';
            }
            cout<<"\n";
        }
        f=1;
        return;
    }
    if(h[p]){
        dfs(p+1);
        return;
    }
    for(int i=0;i<8;i++){
        if(!l[i]&&!zx[p+i]&&!yx[p-i+7]){
            dt[p][i]=1;
            l[i]=1;
            zx[p+i]=1;
            yx[p-i+7]=1;
            dfs(p+1);
            dt[p][i]=0;
            l[i]=0;
            zx[p+i]=0;
            yx[p-i+7]=0;
        }
    }
}
int main(){
    cin>>n;
    for(int i=0;i<n;i++){
        int a,b;
        cin>>a>>b;
        h[a]=1;
        l[b]=1;
        yx[a-b+7]=1;
        zx[a+b]=1;
        dt[a][b]=1;
    }
    dfs(0);
    return 0;
}