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<cassert>
using namespace std;

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

int row[N], col[N], dpos[2 * N - 1], dneg[2 * N - 1];
bool X[N][N];

void initialize() {
	for (int i = 0; i < N; i++) {
		row[i] = FREE;
		col[i] = FREE;
	}
	for (int i = 0; i < 2 * N - 1; i++) {
		dpos[i] = FREE;
		dneg[i] = FREE;
	}
}
void printBoard() {
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (X[i][j]) {
				if (row[i] != j) return;
			}
		}
	}
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			cout << ((row[i] == j) ? "Q" : ".");
		}
		cout << endl;
	}
}
void recursive(int i) {
	if (i == N) {
		printBoard();
		return;
	}
	for (int j = 0; j < N; j++) {
		if (NOT_FREE == col[j] || NOT_FREE == dpos[i + j] || NOT_FREE == dneg[i - j + N - 1]) continue;
		row[i] = j;
		col[j] = dpos[i + j] = dneg[i - j + N - 1] = NOT_FREE;
		recursive(i + 1);
		row[i] = col[j] = dpos[i + j] = dneg[i - j + N - 1] = FREE;
	}
}
int main() {
	initialize();
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) X[i][j] = false;
	}
	int k;
	cin >> k;
	for (int i = 0; i < k; i++) {
		int r, c;
		cin >> r >> c;
		X[r][c] = true;
	}
	recursive(0);
	return 0;
}
