#include <bits/stdc++.h>
using namespace std;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n, k; cin >> n >> k;
  if (2 * k > n + 1) {
    return cout << -1 << '\n', 0;
  }
  int n0 = (n + 1) / 2;
  for (int i = 0; i < n0; ++i) {
    cout << k + i << ' ' << k + 2 * n - n0 + i << ' ' << k + 2 * n + 2 * i << '\n';
  }
  for (int i = n0; i < n; ++i) {
    cout << k + i << ' ' << k + n + i - n0 << ' ' << k + 2 * n + 2 * (i - n0) + 1 << '\n';
  }
}
### Prompt

Your task is to create 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;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n, k; cin >> n >> k;
  if (2 * k > n + 1) {
    return cout << -1 << '\n', 0;
  }
  int n0 = (n + 1) / 2;
  for (int i = 0; i < n0; ++i) {
    cout << k + i << ' ' << k + 2 * n - n0 + i << ' ' << k + 2 * n + 2 * i << '\n';
  }
  for (int i = n0; i < n; ++i) {
    cout << k + i << ' ' << k + n + i - n0 << ' ' << k + 2 * n + 2 * (i - n0) + 1 << '\n';
  }
}
```