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;

class Queen {
public:
	int Q[8][8], R[8], C[8], posit[8 * 2 - 1], negat[2 * 8 - 1]; 

	Queen() {
		for (int i = 0; i < 8; i++) {
			for (int j = 0; j < 8; j++) {
				Q[i][j] = 0;
			}
		}
		for (int i = 0; i < 8; i++) {
			R[i] = 0; 
			C[i] = 0;
		}
		for (int i = 0; i < 2 * 8 - 1; i++) {
			posit[i] = 0;
			negat[i] = 0;
		}
	}

	void recursive(int i) {
		if (i == 8) {
			displayQ();
			return;
		}
		for (int j = 0; j < 8; j++) { 
			if (C[j] || posit[i + j] || negat[i - j + 8 - 1]) { 
				continue;
			}
			R[i] = j; 
			C[j] = posit[i + j] = negat[i - j + 8 - 1] = 1; 
			recursive(i + 1);
			R[i] = C[j] = posit[i + j] = negat[i - j + 8 - 1] = 0; 
		}
	}

	void setQ(int x, int y) {
		Q[x][y] = 1;
	}

	void displayQ() {
		for (int i = 0; i < 8; i++) {
			for (int j = 0; j < 8; j++) {
				if (Q[i][j] && R[i] != j) return; 
			}
		}

		for (int i = 0; i < 8; i++) {
			for (int j = 0; j < 8; j++) {
				cout << ((R[i] == j) ? 'Q' : '.');
			}
			cout << endl;
		}
	}
};

int main() {
	int k;
	cin >> k;
	Queen q = Queen();
	int x, y;
	for (int i = 0; i < k; i++) {
		cin >> x;
		cin >> y;
		q.setQ(x, y);
	}

	q.recursive(0);

	return 0;
}

