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<cstdio>  
#include<algorithm>
#include<stack>
using namespace std;
const int maxx = 8;
// 行列、斜列
int row[maxx], col[maxx];
int dpos[maxx * 2 - 1], dneg[maxx * 2 - 1];

int x[maxx][maxx];

void init() {
	for (int i = 0; i < maxx; i++)
		row[i] = col[i] = 0;
	for (int i = 0; i < maxx * 2 - 1; i++)
		dpos[i] = dneg[i] = 0;
	for (int i = 0; i < maxx; i++)
		for (int j = 0; j < maxx; j++)
			x[i][j] = 0;
}
void print() {

	for (int i = 0; i < maxx; i++) {
		for (int j = 0; j < maxx; j++) {
			if (x[i][j] && row[i] != j)
				return;
		}
	}
	for (int i = 0; i < maxx; i++) {
		for (int j = 0; j < maxx; j++) {
			printf("%c", row[i] == j ? 'Q' : '.');
		}
		printf("\n");
	}
}
void recursive(int i) {
	if (i == maxx) {
		print();
		return;
	}

	for (int j = 0; j < maxx; j++) {
		// 不适合放在这里
		if (col[j] || dpos[i + j] || dneg[i - j + maxx - 1])
			continue;

		// 尝试放下这里
		row[i] = j; col[j] = dpos[i + j] = dneg[i - j + maxx - 1] = 1;
		recursive(i + 1);
		// 恢复刚才的操作继续其他的操作
		row[i] = col[j] = dpos[i + j] = dneg[i - j + maxx - 1] = 0;
	}
}
int main()
{
	int k;
	int r, c;
	scanf("%d", &k);
	for (int i = 0; i < k; i++){
		scanf("%d %d", &r, &c);
		x[r][c] = 1;
	}

	recursive(0);
	return 0;
}
