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>
#include <string.h>
using namespace std;
int eight_queen[8];
int final = 0;

bool check(int row, int value) {
	int data;
	for (int i = 0; i < 8; i++) {
		if (eight_queen[i] != -1) {
			data = eight_queen[i];
			if (value == data) return false;
			if ((value + row) == (data + i)) return false;
			if ((value - row) == (data - i)) return false;
		}
	}
	return true;
}

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

void solution(int row) {
	if (row > 7 && (eight_queen[7] != -1)) {
		final = 1;
		print();
		return;
	}

	if (eight_queen[row] != -1 ) solution(row + 1);
	else {
		for (int index = 0; index < 8; index++) {	
			if (final == 1) return;
			if (check(row, index)) {
				eight_queen[row] = index;
				solution(row + 1);
				eight_queen[row] = -1;
			}
		}
	}
}



int main() {
	int n;
	cin >> n;
	int row, index;
	memset(eight_queen, -1, sizeof(eight_queen));
	for (int i = 0; i < n; i++) {
		cin >> row >> index;
		eight_queen[row] = index;
	}
	solution(0);
	return 0;
}