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...
//lec04
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<stdio.h>
#include<fstream>
#include<stdlib.h>
#include<math.h> 
#include<queue>
#include<string.h>
#include<stack>


using namespace std;

#define N 8
bool board[N][N];
bool col[N],dpos[2*N-1],dneg[2*N-1];
int row[N];


void print_board(){
	for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			if(board[i][j]&&row[i]!=j){
				return;
    		}
    	}
    }
    for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			if(board[i][j]||row[i]==j)cout<<"Q";
      		else cout<<".";
		}
		cout<<endl;
	}
}

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



int main(){

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



