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
#define FREE -1
#define NOT_TREE 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; }
	for(int i=0;i < N;++i){
		for(int j=0;j < N;++j) X[i][j] = false;
	}
}

void print(){
	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 << "\n";
	}
}

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

int main(){
	initialize();
	
	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;
}
