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 initia(){
  for(int i=0; i<N; i++){
    row[i]=FREE,col[i]=FREE;
  }
  for(int i=0; i<2*N-1; i++){
    Dpos[i]=FREE; Dneg[i]=FREE;
  }
}
  void print(){
    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){
      print(); return;
    }

    for(int 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(){
    initia();
      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);

    return 0;
  }

