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 F -1
#define NF 1

int r[N], c[N], dp[2 * N - 1], dn[2 * N - 1];
bool x[N][N];

void print() {
	int i, j;
	for (i = 0; i < N; i++) {
		for (j = 0; j < N; j++) {
			if (x[i][j]) {
				if (r[i] != j) {
					return;
				}
			}
		}
	}
	for (i = 0; i < N; i++) {
		for (j = 0; j < N; j++) {
			cout << ((r[i] == j) ? "Q" : ".");
		}
		cout << endl;
	}
 }

void init() {
	int i;
	for (i = 0; i < N; i++) { r[i] = F, c[i] = F; }
	for (i = 0; i < 2 * N - 1; i++) { dp[i] = F; dn[i] = F; }
}

void recur(int i) {
	int j;
	if (i == N) {
		print();
		return;
	}
	for (j = 0; j < N; j++) {
		if (NF == c[j] || NF == dp[i + j] || NF == dn[i - j + N - 1]) {
			continue;
		}
		r[i] = j;
		c[j] = dp[i + j] = dn[i - j + N - 1] = NF;
		recur(i + 1);
		r[i] = c[j] = dp[i + j] = dn[i - j + N - 1] = F;
	}
}

int main(void) {
	int i,j, k,r,c;
	init();
	for (i = 0; i < N; i++) {
		for (j = 0; j < N; j++) {
			x[i][j] = false;
		}
	}
	cin >> k;
	for (i = 0; i < k; i++) {
		cin >> r >> c;
		x[r][c] = true;
	}
	recur(0);
	return 0;
}
