#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

#define rep(i, N) for (int i = 0; i < (int)N; i++)
const ll MOD = pow(10,9)+7;
const ll LLINF = pow(2,61)-1;
const int INF = pow(2,30)-1;


int main() {
  int N, K; cin >> N >> K;
  if ((N+1)/2<K) { cout << -1 << endl; return 0;}
  int l = (N+1)/2;
  rep(i,l) {
    int a = K+i;
    int b = K+(2*N-l)+i;
    int c = K+2*N+i*2;
    cout << a << " " << b << " " << c << endl;
  }
  rep(i,N-l) {
    int a = K+l+i;
    int b = K+N+i;
    int c = K+2*N+i*2+1;
    cout << a << " " << b << " " << c << endl;
  }
  return 0;
}
### Prompt

Create a solution in CPP for 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;
typedef long long ll;

#define rep(i, N) for (int i = 0; i < (int)N; i++)
const ll MOD = pow(10,9)+7;
const ll LLINF = pow(2,61)-1;
const int INF = pow(2,30)-1;


int main() {
  int N, K; cin >> N >> K;
  if ((N+1)/2<K) { cout << -1 << endl; return 0;}
  int l = (N+1)/2;
  rep(i,l) {
    int a = K+i;
    int b = K+(2*N-l)+i;
    int c = K+2*N+i*2;
    cout << a << " " << b << " " << c << endl;
  }
  rep(i,N-l) {
    int a = K+l+i;
    int b = K+N+i;
    int c = K+2*N+i*2+1;
    cout << a << " " << b << " " << c << endl;
  }
  return 0;
}
```