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>
using namespace std;

int row[8];
bool col[8] , dpos[15] , dneg[15];

void printBoard(){
  for(int i=0; i<8; i++){
    for(int j=0; j<8; j++){
      if(row[i] == j)cout << 'Q';
      else cout << '.';
    }
    cout << endl;
  }

}


bool putQueen(int n){
  if(n==8){
    printBoard();
    return true;
  }

  if( row[n]!=-1 ){
    return putQueen(n+1);
  }

  
  for(int i=0; i<8; i++){
    if(col[i] || dpos[n+i] || dneg[n-i+7] )continue;

    row[n] = i;
    col[i] = dpos[n+i] = dneg[n-i+7] = true;
    if( putQueen(n+1) )return true;

    col[i] = false;
    dpos[n+i] =  false;
    dneg[n-i+7] = false;
  }

  row[n] = -1;
  return false;
}





int main(){

  int k,x,y;


  //init
  for(int i=0; i<8; i++){ row[i]=-1; col[i]=false; }
  for(int i=0; i<15; i++){ dpos[i] = dneg[i] = false; }
  
  //input
  cin >> k;
  for(int i=0; i<k; i++){
    cin >> y >> x;
    row[y] = x;
    col[x] = true;
    dpos[y+x] = dneg[y-x+7] = true;
    
  }

  //solve
  putQueen(0);

  return 0;
}

