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;


int main()
{
	int k;
	cin>> k;
	vector<int> p(8), q(8), order(8);
	for(int i=0;i<k;i++){
		cin>> p[i] >> q[i];
	}
	
	for(int i=0;i<8;i++){
		order[i] = i;
	}
	do{
		bool flag = false;
		for(int i=0;i<k;i++){
			if(order[p[i]]!=q[i]){
				flag = true;
				break;
			}
		}
		if(flag) continue;
		for(int i=1;i<8;i++){
			for(int j=0;j<8;j++){
				if(i != j && abs(i-j) == abs(order[i]-order[j])){
					flag = true;
					break;
				}
			}
		}
		if(flag) continue;
		if(!flag) break;
	}while(next_permutation(order.begin(), order.end()));
	
	for(int i=0;i<8;i++){
		for(int j=0;j<8;j++){
			if(j<7&&order[i]==j){
				cout<<"Q";
			}else if(j<7&&order[i]!=j){
				cout<<".";
			}
			if(j==7&&order[i]==j){
				cout<<"Q"<<endl;
			}else if(j==7&&order[i]!=j){
				cout<<"."<<endl;
			}
		}
	}
}

