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>
#include<cassert>
using namespace std;

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

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

bool X[N][N];

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

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++)
		{
			cout<<((row[i]==j)?"Q":".");
		}
		cout<<endl;
	}
}

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

int main()
{
	init();
	
	for(int i=0;i<N;i++)
	{
		for(int j=0;j<N;j++)X[i][j]=false;
	} 
	
	int k;cin>>k;

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

}
