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;

#define N 8

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

bool X[N][N];

void print() {
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			cout << (row[i] == j ? "Q" : ".");
		}
		cout << endl;
	}
}

void solve(int i) {
	if (i == N) {
		print();
		return;
	}
	if (row[i] != -1) {
		solve(i + 1);
		return;
	}
	for (int j = 0; j < N; ++j) {
		if (col[j] != -1 || dpos[i + j] != -1 || dneg[i - j + N - 1] != -1) continue;
		row[i] = j; col[j] = dpos[i + j] = dneg[i - j + N - 1] = 1;
		solve(i + 1);
		row[i] = col[j] = dpos[i + j] = dneg[i - j + N - 1] = -1;
	}
}

int main() {
	for (int i = 0; i < N; ++i) row[i] = col[i] = -1;
	for (int i = 0; i < 2 * N - 1; ++i) dpos[i] = dneg[i] = -1;

	int k; cin >> k;
	while (k--) {
		int r, c; cin >> r >> c;
		X[r][c] = true;
		row[r] = c; col[c] = dpos[r + c] = dneg[r - c + N - 1] = 1;
	}

	solve(0);

	return 0;
}
