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;

int f[8][8];
int dx[] = { 1,1,0,-1,-1,-1,0,1 };
int dy[] = { 0,-1,-1,-1,0,1,1,1 };

int queen(int y) {
	if (y == 8)
		return 1;
	for (int x = 0; x < 8; ++x) {
		if (f[y][x]) {
			return queen(y + 1);
		}
		int flag = 0;
		for (int k = 0; k < 8; ++k) {
			int nx = x, ny = y;
			while (!flag) {
				nx += dx[k];
				ny += dy[k];
				if (nx < 0 || nx > 7 || ny < 0 || ny > 7)
					break;
				if (f[ny][nx])
					flag = 1;
			}
			if (flag)
				break;
		}
		if (!flag) {
			f[y][x] = 1;
			if (queen(y + 1))
				return 1;
			f[y][x] = 0;
		}
	}
	return 0;
}

int main() {
	int k, r, c;
	cin >> k;
	while (k--) {
		cin >> r >> c;
		f[r][c] = 1;
	}
	queen(0);
	for (int i = 0; i < 8; ++i) {
		for (int j = 0; j < 8; ++j) {
			cout << (f[i][j] ? 'Q' : '.');
		}
		cout << endl;
	}

	return 0;
}
