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

void printBoard(){
	for(int i=0; i<N; i++){
		for(int j=0; j<N; j++){
			if(j==row[i]){ cout << "Q"; }
			else{ cout << "."; }
		} 
		cout<<endl;
	}
}

void putQueen(int i){
	if(i == N){ printBoard(); return; }

	if(reservedQ[i] == 1){ putQueen(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;
		
		putQueen(i+1);
		
		col[j] = dpos[i+j] = dneg[i-j+N-1] = 0;
	}
}

int main(){
	int k, r, c;
	 
	cin >> k;
	for(int i=0; i<k; i++){
		cin >> r >> c;
		row[r] = c;
		col[c] = dpos[r+c] = dneg[r-c+N-1] = 1;
		reservedQ[r] = 1;
	}
	
	putQueen(0);
	
	return 0;
}
