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 <algorithm>
#include <iostream>
#include <math.h>
#include <stdio.h>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
using namespace std;
typedef long long ll;

int main()
{
	int k;
	cin>>k;
	map<int,int> bc;
	rep(i,k){
		int x,y;
		cin>>x>>y;
		bc.insert(make_pair(x,y));
	}
	int queen[8];
	rep(i,8){
		queen[i] = i;
	}
	do{
		bool canMake = true;
		for(auto iter = bc.begin(); iter != bc.end(); iter++){
			int x= iter->first;
			int y= iter->second;
			if( queen[x] != y){
				canMake = false;
				break;
			}
		}
		for(int i = 0 ; i < 8; i++){
			for(int j = i + 1; j < 8; j++){
				if( queen[j] == queen[i] + (j-i) || queen[j] == queen[i] - (j-i)){
					canMake = false;
					break;
				}
			}
		}
		if(canMake == true){
			break;
		}
	}while(next_permutation(queen,queen+8));

	for(int i = 0; i < 8;i++){
		for(int j = 0; j < 8; j++){
			if( queen[i] == j){
				cout<<"Q";
			}else{
				cout<<".";
			}
		}
		cout<<endl;
	}
	return 0;
}


