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<bits/stdc++.h>
#define N 8
#define FREE -1
#define NFREE 1
using namespace std;

vector<int> row(N),col(N),dpos(2*N-1),dneg(2*N-1);
vector<vector<bool>> data(N,vector<bool>(N));

void init(){
  for(int i=0;i<N;i++){
    row.at(i) = FREE;
    col.at(i) = FREE;
  }
  for(int i=0;i<2*N-1;i++){
    dpos.at(i) = FREE;
    dneg.at(i) = FREE;
  }
}

void printdata(){
	for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			if(data.at(i).at(j)){
				if(row.at(i) != j) return; 
			}
		}
	}
	
	for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			if(row.at(i) == j) cout << "Q";
			else cout << ".";
		}
	cout << endl;
	}
}

void rec(int i){
	if(i == N){
		printdata();
		return;
	}
	for(int j=0;j<N;j++){
		if(col.at(j) == 1 || dpos.at(i+j) == 1 || dneg.at(i-j+N-1) == 1) continue;
		row.at(i) = j;
		col.at(j) = dpos.at(i+j) = dneg.at(i-j+N-1) = 1;
		rec(i+1);
		row.at(i) = col.at(j) = dpos.at(i+j) = dneg.at(i-j+N-1) = 0;
	}
}

int main(){
	init();
	for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			data.at(i).at(j) = 0;
		}
	}
	int k,r,c;
	cin >> k;
	for(int i=0;i<k;i++){
		cin >> r >> c;
		data.at(r).at(c) = 1;
	}
	rec(0);
}
