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 <stdio.h>
#include <algorithm>
using namespace std;
#include <string.h>
int vis[20][20];
int book[10][10];
int map[10][10];
int flag,n;
void dfs(int x)
{
	if(flag) return ;
	if(x>=8)
	{
		for(int i=0;i<8;i++)
		{
			for(int j=0;j<8;j++)
			if(map[i][j]) printf("Q");
			else printf(".");
			printf("\n");
		}
		flag=1;
		return ;
	} 
	for(int i=0;i<8;i++)
	{
	  if(map[x][i]) 
        dfs(x+1);
	  else if(!vis[0][i]&&!vis[1][x+i]&&!vis[2][x-i+n])
	  	  {
	  		vis[0][i]=vis[1][x+i]=vis[2][x-i+n]=1;
	  		map[x][i]=1;
	  		dfs(x+1);
	  		map[x][i]=0;
	  		vis[0][i]=vis[1][x+i]=vis[2][x-i+n]=0;
		 } 
	}
	return;
}
int main()
{
	
	while(~scanf("%d",&n))
	{
	   int x,y;flag=0;
	   memset(book,0,sizeof(map));
	   memset(book,0,sizeof(book));
	   memset(vis,0,sizeof(vis));
	   for(int i=1;i<=n;i++)
	   {
	   	scanf("%d %d",&x,&y);
	   	map[x][y]=1;
	   vis[0][y]=vis[1][x+y]=vis[2][x-y+n]=1;
	   }
	   dfs(0);
	}
	return 0;
}
