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>
using namespace std;

#define N 8
#define FREE -1
#define NOT_FREE 1

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

void printBoard()
{
	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++ ) {
			char a = '.';
			if( row[i] == j ) a = 'Q';
			cout << a;
		}
		cout << endl;
	}
}

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

int main()
{
	int k;
	cin >> k;

	for( int i = 0; i < k; i++ ) {
		int r, c;
		cin >> r >> c;
		X[r][c] = 1;
	}
	dfs( 0 );
}

