#include <bits/stdc++.h>

using namespace std;
using int64 = long long;

constexpr int DEBUG = 0;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n, k;
  cin >> n >> k;
  if (n < 2 * k - 1) {
    cout << -1 << endl;
    return 0;
  }

  vector<int> xs(n);
  vector<int> ys(n);
  vector<int> zs(n);

  for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      xs[i] = k + i / 2;
      ys[i] = k + n + (n / 2) + i / 2;
    } else {
      xs[i] = k + ((n + 1) / 2) + i / 2;
      ys[i] = k + n + i / 2;
    }
    zs[i] = k + 2 * n + i;
  }

  for (int i = 0; i < n; i++) {
    cout << xs[i] << " " << ys[i] << " " << zs[i] << "\n";
  }
}
### Prompt

Please formulate 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 int64 = long long;

constexpr int DEBUG = 0;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n, k;
  cin >> n >> k;
  if (n < 2 * k - 1) {
    cout << -1 << endl;
    return 0;
  }

  vector<int> xs(n);
  vector<int> ys(n);
  vector<int> zs(n);

  for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      xs[i] = k + i / 2;
      ys[i] = k + n + (n / 2) + i / 2;
    } else {
      xs[i] = k + ((n + 1) / 2) + i / 2;
      ys[i] = k + n + i / 2;
    }
    zs[i] = k + 2 * n + i;
  }

  for (int i = 0; i < n; i++) {
    cout << xs[i] << " " << ys[i] << " " << zs[i] << "\n";
  }
}
```