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>
using namespace std;

int queen[9];

void output(){
    for(int i=1;i<=8;i++){
        int t=queen[i];t--;
        while(t--) cout<<".";
        cout<<"Q";
        t=8-queen[i];
        while(t--) cout<<".";
        cout<<endl;
    }
}
void dfs(int cur){
    int flag,i,j;

    if(cur==9){ 
        output();
    }
    else if(queen[cur]){
        dfs(cur+1);
    }
    else{
        for(i=1;i<=8;i++){
            flag=1;
            for(j=1;j<=8;j++){
                if(queen[j]&&(queen[j]==i || i+cur==queen[j]+j || i-cur==queen[j]-j)){
                    flag=0;
                    break;
                }
            }
            if(flag){
                queen[cur]=i;
                dfs(cur+1);
                queen[cur]=0;
            }
        }
    }
}
int main(){
    int n;cin>>n;
    while(n--){
        int x,y;
        cin>>x>>y;
        queen[x+1]=y+1;
    }
    dfs(1);
}
