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 <iostream>
using namespace std;
#define FREE 100
#define NOT_FREE 200
#define N 8

int col[20],row[20],dpos[20],dneg[20];
int ans[20];

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

void putQueen(int i){
	//cout << i << endl;
	if(i == 8){
		printBoard();
		return;
	}

	if(ans[i] == FREE){
		for(int j = 0;j < N;j++){
			if(col[j] == NOT_FREE || dpos[i+j] == NOT_FREE || dneg[i-j+N-1] == NOT_FREE){
				continue;
			}
			row[i] = j;
			col[j] = dpos[i+j] = dneg[i-j+N-1] = NOT_FREE;
			putQueen(i+1);
			col[j] = dpos[i+j] = dneg[i-j+N-1] = FREE;
		}
	}
	else{
		int j = ans[i];
		if(col[j] == NOT_FREE || dpos[i+j] == NOT_FREE || dneg[i-j+N-1] == NOT_FREE){
		}
		else{
			row[i] = j;
			col[j] = dpos[i+j] = dneg[i-j+N-1] = NOT_FREE;
			putQueen(i+1);
			col[j] = dpos[i+j] = dneg[i-j+N-1] = FREE;
		}
	}
}

int main(){
	for(int i = 0;i < 20;i++){
		col[i] = dpos[i] = dneg[i] = ans[i] = FREE;
	}

	int k,r,c;
	cin >> k;
	for(int i = 0;i < k;i++){
		cin >> r >> c;
		ans[r] = c;
		//col[c] = dpos[r+c] = dneg[r-c+N-1] = NOT_FREE;
	}

	putQueen(0);

	return 0;
}
