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 f=0;
	int k;
	cin >> k;
	vector<int>x(k),y(k);
	for(int i=0; i<k; i++) cin >> x.at(i) >> y.at(i);
	vector<int>a(8),b(8);
	for(int i=0; i<8; i++) a.at(i)=i, b.at(i)=i;
	do{
		do{
			int cnt = 0;
			for(int j=0; j<k; j++){
				for(int i=0; i<8; i++){
					if((x.at(j)==a.at(i) && y.at(j)==b.at(i))) cnt++;
				}
			}
			if(cnt<k) continue;
			bool flg = 1;
			for(int i=0; i<7; i++){
				for(int j=i+1; j<8; j++){
					if((a.at(i)-a.at(j)==b.at(i)-b.at(j))||(a.at(i)-a.at(j)==b.at(j)-b.at(i)))flg=0;
				}
			}
			if(flg){
				for(int i=0; i<8; i++){
					for(int j=0; j<8; j++){
						bool flg2=0;
						for(int k=0; k<8; k++){
							if(a.at(k)==i && b.at(k)==j) flg2=1;
						}
						if(flg2) cout << 'Q';
						else cout << '.';
					}
					cout << endl;
				} 
				return 0;
			}
		}while(next_permutation(b.begin(),b.end()));
	}while(next_permutation(a.begin(),a.end()));
	
}
