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;

const int N = 8;

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

enum {
	FREE = 20,NOT_FREE
};

bool Graph[N][N];

void initialize() {
	for (int i=0;i<N;i++) {
		for (int j=0;j<N;j++) {
			Graph[i][j] = false;
		}
	}
	
	for (int i=0;i<N;i++) {
		row[i] = col[i] = FREE;
	}
	
	for (int i=0;i<2*N-1;i++) {
		dpos[i] = dneg[i] = FREE;
	}
}

void printQ() {
	for (int i=0;i<N;i++) {
		for (int j=0;j<N;j++) {
			if (Graph[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 solve(int i) {
	if (i == N) {
		printQ();
		return;
	}
	
	for (int j=0;j<N;j++) {
		if (row[i] == FREE && col[j] == FREE && dpos[i+j] == FREE && dneg[i-j+N-1] == FREE) {
			row[i] = j;
			col[j] = dpos[i+j] = dneg[i-j+N-1] = NOT_FREE;
			solve(i+1);
			row[i] = col[j] = dpos[i+j] = dneg[i-j+N-1] = FREE;
 		}
	}
}

int main() {
	initialize();
	int k;
	cin >> k;
	while (k--) {
		int x,y;
		cin >> x >> y;
		Graph[x][y] = true;
	}
	solve(0);
	return 0;
}
