#include<bits/stdc++.h>
using namespace std;
int main(){
  int64_t N,K;cin>>N>>K;
  if(1<2*K-N)
    cout<<-1<<endl;
  else{
  if(N%2==1){
  int64_t k=(N-1)/2;
    for(int64_t i=0;i<=k;i++)
      cout<<K+2*i<<" "<<K+N+k-i<<" "<<K+2*N+i<<endl;
    for(int64_t i=0;i<k;i++)
      cout<<K+2*i+1<<" "<<K+2*N-1-i<<" "<<K+2*N+i+k+1<<endl;
  }else{
  int64_t k=N/2;
    for(int64_t i=0;i<=k-1;i++)
      cout<<K+2*i<<" "<<K+N+k-i<<" "<<K+2*N+i<<endl;
    for(int64_t i=0;i<k-1;i++)
      cout<<K+2*i+1<<" "<<K+N+N+i-2*i-1<<" "<<K+2*N-k+N+i<<endl;
    cout<<K+N-1<<" "<<K+N<<" "<<K+3*N-1<<endl;
  }
  }
  return 0;
}
### Prompt

Generate 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(){
  int64_t N,K;cin>>N>>K;
  if(1<2*K-N)
    cout<<-1<<endl;
  else{
  if(N%2==1){
  int64_t k=(N-1)/2;
    for(int64_t i=0;i<=k;i++)
      cout<<K+2*i<<" "<<K+N+k-i<<" "<<K+2*N+i<<endl;
    for(int64_t i=0;i<k;i++)
      cout<<K+2*i+1<<" "<<K+2*N-1-i<<" "<<K+2*N+i+k+1<<endl;
  }else{
  int64_t k=N/2;
    for(int64_t i=0;i<=k-1;i++)
      cout<<K+2*i<<" "<<K+N+k-i<<" "<<K+2*N+i<<endl;
    for(int64_t i=0;i<k-1;i++)
      cout<<K+2*i+1<<" "<<K+N+N+i-2*i-1<<" "<<K+2*N-k+N+i<<endl;
    cout<<K+N-1<<" "<<K+N<<" "<<K+3*N-1<<endl;
  }
  }
  return 0;
}
```