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 initialize()
{
  int i;

  for(i=0;i<N;i++)
    {
      row[i]=FREE,col[i]=FREE;
    }
  for(i=0;i<N;i++)
    {
      dpos[i]=FREE,dneg[i]=FREE;
    }
}

void printBoard()
{
  int i,j;

  for(i=0;i<N;i++)
    {
      for(j=0;j<N;j++)
	{
	  if(X[i][j])
	    {
	      if(row[i]!=j) return;
	    }
	}
    }
  for(i=0;i<N;i++)
    {
      for(j=0;j<N;j++)
	{
	  cout<<((row[i]==j)?"Q":".");
	}
      cout<<endl;
    }
}


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

  for(j=0;j<N;j++)
    {
      if(NOT_FREE==col[j] || NOT_FREE == dpos[i+j] || NOT_FREE == dneg[i-j+N-1])
	{
	  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()
{
  initialize();

  int i,j,k,r,c;

  for(i=0;i<N;i++)
    {
      for(j=0;j<N;j++)
	{
	  X[i][j]=false;
	}
    }

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

  recursive(0);

  return 0;
}

