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>
#define MAX 8
using namespace std;

bool c[MAX],x[MAX*2-1],y[MAX*2-1];
int r[MAX];

void init(){
  for(int i=0;i<MAX;i++){
    r[i]=-1;
    c[i]=false;
  }
  for(int i=0;i<MAX*2-1;i++){
    x[i]=false;
    y[i]=false;
  }
}

void print(){
  for(int i=0;i<MAX;i++){
    for(int j=0;j<MAX;j++){
      if(r[i]==j)cout<<'Q';
      else cout<<'.';
    }
    cout<<endl;
  }
}

void rec(int n){
  if(n==MAX)print();
  else if(r[n]!=-1)rec(n+1);
  else {
    for(int i=0;i<MAX;i++){
      if(!c[i] && !x[MAX-1-n+i] && !y[n+i]){
	r[n]=i;
	c[i]=true;
	x[MAX-1-n+i]=true;
	y[n+i]=true;
	rec(n+1);
	r[n]=-1;
	c[i]=false;
	x[MAX-1-n+i]=false;
	y[n+i]=false;
      }
    }
  }
}

int main()
{
  int n,a,b;

  init();

  cin>>n;
  for(int i=0;i<n;i++){
    cin>>a>>b;
    r[a]=b;
    c[b]=true;
    x[MAX-1-a+b]=true;
    y[a+b]=true;
  }

  rec(0);

  return 0;
}