#include<iostream>
#include<map>
#include<set>
using namespace std;
#define df 0
#define MOD 998244353
typedef long int li;

void print(const map<int,li>& mp){
  for(const auto& x:mp){
    printf("(%d %ld)",x.first,x.second);
  }
  printf("\n");
}

int main(){
  if(df) printf("*debug mode*\n");

  li n,k; cin >>n >>k;
  if(2*k-1>n){
    printf("-1");
    return 0;
  }
  li l=k+2*n;
  for(int i=0;i<k;i++){
    printf("%ld %ld %ld\n",k+i,l-k+i,l+2*i);
  }
  for(int i=0;i<k-1;i++){
    printf("%ld %ld %ld\n",2*k+i,l-2*k+i+1,l+2*i+1);
  }
  for(int i=0;2*k-1+i<n;i++){
    printf("%ld %ld %ld\n",3*k-1+i,l-2*k-i,l+2*k-1+i);
  }
}

/// confirm df==0 ///

### Prompt

Please provide a CPP coded solution to the problem described below:

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<iostream>
#include<map>
#include<set>
using namespace std;
#define df 0
#define MOD 998244353
typedef long int li;

void print(const map<int,li>& mp){
  for(const auto& x:mp){
    printf("(%d %ld)",x.first,x.second);
  }
  printf("\n");
}

int main(){
  if(df) printf("*debug mode*\n");

  li n,k; cin >>n >>k;
  if(2*k-1>n){
    printf("-1");
    return 0;
  }
  li l=k+2*n;
  for(int i=0;i<k;i++){
    printf("%ld %ld %ld\n",k+i,l-k+i,l+2*i);
  }
  for(int i=0;i<k-1;i++){
    printf("%ld %ld %ld\n",2*k+i,l-2*k+i+1,l+2*i+1);
  }
  for(int i=0;2*k-1+i<n;i++){
    printf("%ld %ld %ld\n",3*k-1+i,l-2*k-i,l+2*k-1+i);
  }
}

/// confirm df==0 ///

```