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;
#define N 8

vector<int> row(N,-1),col(N,-1),dpos(2*N-1,-1),dneg(2*N-1);
vector<vector<int>> X(N,vector<int>(N,0));

void printBoard(){
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            if(X[i][j]){
                if(row[i]!=j) return;
            }
        }
    }
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            cout<<(row[i]==j?'Q':'.');
        }
        cout<<endl;
    }
}

void recursive(int i){
    if(i==N){
        printBoard();
    }
    for(int j=0;j<N;j++){
        if(col[j]==1||dpos[i+j]==1||dneg[i-j+N-1]==1) continue;
        row[i]=j;
        col[j]=dpos[i+j]=dneg[i-j+N-1]=1;
        recursive(i+1);
        row[i]=col[j]=dpos[i+j]=dneg[i-j+N-1]=-1;
    }
}

int main(){
    int k;
    cin>>k;
    int a,b;
    for(int i=0;i<k;i++){
        cin>>a>>b;
        X[a][b]=1;
    }

    recursive(0);
}
