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;

#define FREE -1
#define NOT_FREE 0
#define N 8
 
 
int row[N],col[N],pos[2*N-1],neg[2*N-1];
int sol[N];

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

void printBoard(){
  int i,j;

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

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

int main(){
  int k,r,c;
  
  init();
  cin >> k;
  
  for (int i = 0; i < k; i++){
    cin >> r >> c;
    sol[r]=c; 
  }
  nqueen(0);
  return 0;
}


