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...
#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stack>
using namespace std;
const int num = 8;
int tf[num][num], col[num], ld[num*2-1], rd[num*2-1], row[num];
inline void print() {
	for (int i = 0; i < num; i++) {
		for (int j = 0; j < num; j++) {
			if (tf[i][j])
				if (row[i] != j) return;
		}
	}
	for (int i = 0; i < num; i++) {
		for (int j = 0; j < num; j++) {
			cout << ((row[i] == j) ? "Q" : ".");
		}
		cout << endl;
	}
}
inline void queen(int i) {
	if (i == num) {
		print();
		return;
	}
	else {
		for (int j = 0; j < num; j++) {
			if (!col[j] || !ld[i + j] || !rd[i - j + num - 1]) continue;
			row[i] = j;
			col[j] = ld[i + j] = rd[i - j + num - 1] = 0;
			queen(i + 1);
			col[j] = ld[i + j] = rd[i - j + num - 1] = 1;
			row[i] = -1;
		}
	}
}
int main() {
	for (int i = 0; i < num; i++) {
		for (int j = 0; j < num; j++) tf[i][j] = 0;
		col[i] = 1; row[i] = -1;
	}
	for (int i = 0; i < num * 2 - 1; i++) ld[i] = rd[i] = 1;
	int n, x, y;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> x >> y;
		tf[x][y] = 1;
	}
	queen(0);
	return 0;
}