#include<bits/stdc++.h>
using namespace std;

int main(){
  long int n,k,i;
  cin >> n >> k;
  if(n<k*2-1){
    cout << -1 << endl;
  }else if(n%2==1){
    for(i=0;i<=(n-1)/2;i++){
      cout << k+2*i << ' ' << k+(3*n-1)/2-i << ' ' << k+2*n+i << endl;
    }
    for(i=0;i<=(n-3)/2;i++){
      cout << k+2*i+1 << ' ' << k+2*n-1-i << ' ' << k+(5*n+1)/2+i << endl;
    }
  }else{
    for(i=0;i<=n/2-1;i++){
      cout << k+2*i << ' ' << k+3*n/2-1-i << ' ' << k+2*n+i << endl;
    }
    for(i=0;i<=n/2-1;i++){
      cout << k+2*i+1 << ' ' << k+2*n-1-i << ' ' << k+5*n/2+i << endl;
    }
  }
}
### Prompt

Your task is to create a CPP solution to the following problem:

Given are positive integers N and K.

Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.

* For every integer i from 1 to N, a_i + b_i \leq c_i holds.



If the answer is yes, construct one such partition.

Constraints

* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9

Input

Input is given from Standard Input in the following format:


N K


Output

If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:


a_1 b_1 c_1
:
a_N b_N c_N

Output

If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:


a_1 b_1 c_1
:
a_N b_N c_N

Examples

Input

1 1


Output

1 2 3


Input

3 3


Output

-1

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

int main(){
  long int n,k,i;
  cin >> n >> k;
  if(n<k*2-1){
    cout << -1 << endl;
  }else if(n%2==1){
    for(i=0;i<=(n-1)/2;i++){
      cout << k+2*i << ' ' << k+(3*n-1)/2-i << ' ' << k+2*n+i << endl;
    }
    for(i=0;i<=(n-3)/2;i++){
      cout << k+2*i+1 << ' ' << k+2*n-1-i << ' ' << k+(5*n+1)/2+i << endl;
    }
  }else{
    for(i=0;i<=n/2-1;i++){
      cout << k+2*i << ' ' << k+3*n/2-1-i << ' ' << k+2*n+i << endl;
    }
    for(i=0;i<=n/2-1;i++){
      cout << k+2*i+1 << ' ' << k+2*n-1-i << ' ' << k+5*n/2+i << endl;
    }
  }
}
```