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 <bits/stdc++.h>
using namespace std;

bool isOK(vector<int> v) {
	for (int i = 0; i < 8; i++) {
		for (int j = i + 1; j < 8; j++) {
			if (abs(v[j] - v[i]) == j - i) {
				return false;
			}
		}
	}
	return true;
}

int main()
{
	int k;
	cin >> k;
	vector<int> x(k), y(k);
	for (int i = 0; i < k; i++) {
		cin >> x[i] >> y[i];
	}
	vector<int> v = { 0, 1, 2, 3, 4, 5, 6, 7 };
	do {
		bool OK = true;
		for (int i = 0; i < k; i++) {
			if (v[x[i]] != y[i]) {
				OK = false;
			}
		}
		if (OK && isOK(v)) {
			break;
		}
	} while (next_permutation(v.begin(), v.end()));
	vector<string> res(8, "........");
	for (int i = 0; i < 8; i++) {
		res[i][v[i]] = 'Q';
	}
	for (int i = 0; i < 8; i++) {
		cout << res[i] << endl;
	}
	return 0;
}