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 <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;

#define N 8
int a[N], b[2 * N - 1], c[2 * N - 1], x[N];
int n;
int xp[N], yp[N];

void found() {
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (x[i] == j) cout << "Q";
			else           cout << ".";
		}
		cout << endl;
	}
}

void try_(int i) {
	for (int j = 0; j < N; j++) {
		if (a[j] && b[i+j] && c[i-j+N-1]) {
			x[i] = j;
			if (i < N -1) {
				a[j] = b[i+j] = c[i-j+N-1] = 0;
				try_(i + 1);
				a[j] = b[i+j] = c[i-j+N-1] = 1;
			}
			else {
				bool flag = true;
				for (int k = 0; k < n; k++)
					if (x[xp[k]] != yp[k])
						flag = false;
				if (flag)
					found();
			}

		}
	}
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> xp[i] >> yp[i];

	fill_n(a, N, 1);
	fill_n(b, 2 * N - 1, 1);
	fill_n(c, 2 * N - 1, 1);
	try_(0);
	return 0;
}