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 <cstdio>
#include <iostream>

using namespace std;

#define N 8
#define FREE -1
#define NOT_FREE 1

int row[N], col[N];
int lowLeft[N * 2 - 1], lowRight[N * 2 - 1];
bool X[N][N];

void printBoard() {
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			if (X[i][j] && row[i] != j)
				return;
		}
	}
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			if (row[i] == j)
				cout << "Q";
			else
				cout << ".";
		}
		cout << endl;
	}
}

void backTrack(int i) {
	if (i == N) {
		printBoard();
		return;
	}
	for (int j = 0; j < N; ++j) {
		if (col[j] == NOT_FREE || lowLeft[i + j] == NOT_FREE || lowRight[i - j + N - 1] == NOT_FREE)
			continue;
		row[i] = j;
		col[j] = lowLeft[i + j] = lowRight[i - j + N - 1] = NOT_FREE;
		backTrack(i + 1);
		row[i] = col[j] = lowLeft[i + j] = lowRight[i - j + N - 1] = FREE;
	}
}

int main() {
	for (int i = 0; i < N; ++i) {
		row[i] = col[i] = FREE;
	}
	for (int i = 0; i < N * 2 - 1; ++i) {
		lowLeft[i] = lowRight[i] = FREE;
	}
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			X[i][j] = false;
		}
	}

	int k;
	scanf("%d", &k);
	for (int i = 0; i < k; ++i) {
		int r, c;
		scanf("%d %d", &r, &c);
		X[r][c] = true;
	}
	backTrack(0);

	return 0;
}