#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;

int main(){
  int n,k;
  cin >> n >> k;
  if(2*k-1>n){
    cout << -1 << endl;
    return 0;
  }
  vector<int> a(n),b(n),c(n);
  for(int i=0;i<n;i+=2){
    a[i]=k; k++;
  }
  for(int i=1;i<n;i+=2){
    a[i]=k; k++;
  }
  for(int i=1;i<n;i+=2){
    b[i]=k; k++;
  }
  for(int i=0;i<n;i+=2){
    b[i]=k; k++;
  }
  for(int i=0;i<n;i++){
    c[i]=k; k++;
  }
  for(int i=0;i<n;i++){
    cout << a[i] << ' ' << b[i] << ' ' << c[i] << endl;
  }
}
### 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 <vector>
#include <algorithm>
using namespace std;
typedef long long ll;

int main(){
  int n,k;
  cin >> n >> k;
  if(2*k-1>n){
    cout << -1 << endl;
    return 0;
  }
  vector<int> a(n),b(n),c(n);
  for(int i=0;i<n;i+=2){
    a[i]=k; k++;
  }
  for(int i=1;i<n;i+=2){
    a[i]=k; k++;
  }
  for(int i=1;i<n;i+=2){
    b[i]=k; k++;
  }
  for(int i=0;i<n;i+=2){
    b[i]=k; k++;
  }
  for(int i=0;i<n;i++){
    c[i]=k; k++;
  }
  for(int i=0;i<n;i++){
    cout << a[i] << ' ' << b[i] << ' ' << c[i] << endl;
  }
}
```