#include <iostream>
#include <vector>
using namespace std;
int N, K;
int main() {
  cin >> N >> K;
  vector<int> a, b;
  bool ok = true;
  for(int i = 0; i < (N+1)/2; ++i) {
    ok &= K+i*2 + K+N+(N+1)/2-1-i <= K+N*2+i;
    a.push_back(K+i*2);
    b.push_back(K+N+(N+1)/2-1-i);
  }
  for(int i = 0; i < N/2; ++i) {
    ok &= K+i*2+1 + K+N*2-1-i <= K+N*2+(N+1)/2+i;
    a.push_back(K+i*2+1);
    b.push_back(K+N*2-1-i);
  }
  if(ok)
  for(int i = 0; i < N; ++i)
    cout << a[i] << " " << b[i] << " " << i+K+N*2 << '\n';
  else cout << -1;
  return 0;
}
### Prompt

Please create a solution in cpp 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 <iostream>
#include <vector>
using namespace std;
int N, K;
int main() {
  cin >> N >> K;
  vector<int> a, b;
  bool ok = true;
  for(int i = 0; i < (N+1)/2; ++i) {
    ok &= K+i*2 + K+N+(N+1)/2-1-i <= K+N*2+i;
    a.push_back(K+i*2);
    b.push_back(K+N+(N+1)/2-1-i);
  }
  for(int i = 0; i < N/2; ++i) {
    ok &= K+i*2+1 + K+N*2-1-i <= K+N*2+(N+1)/2+i;
    a.push_back(K+i*2+1);
    b.push_back(K+N*2-1-i);
  }
  if(ok)
  for(int i = 0; i < N; ++i)
    cout << a[i] << " " << b[i] << " " << i+K+N*2 << '\n';
  else cout << -1;
  return 0;
}
```