#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

ll N, K;

int main() {
  cin >> N >> K;
  if (2 * K - 1 > N)
    cout << -1 << endl;
  else {
    ll L = (N + 1) / 2;
    if (N == 2 * L - 1) {
      for (ll i = 0; i < L; ++i) {
        cout << K + 2 * i << " " << K + 3 * L - 2 - i << " "
             << K + 4 * L - 2 + i << endl;
      }
      for (ll i = 0; i < L - 1; ++i) {
        cout << K + 1 + 2 * i << " " << K + 4 * L - 3 - i << " "
             << K + 5 * L - 2 + i << endl;
      }
    } else {
      for (ll i = 0; i < L; ++i) {
        cout << K + 2 * i << " " << K + (i == 0 ? 2 : 3) * L - i << " "
             << K + 4 * L + i << endl;
      }
      for (ll i = 0; i < L; ++i) {
        cout << K + 1 + 2 * i << " " << K + 4 * L - 1 - i << " "
             << K + 5 * L + i << endl;
      }
    }
  }
  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 <bits/stdc++.h>
using namespace std;

typedef long long ll;

ll N, K;

int main() {
  cin >> N >> K;
  if (2 * K - 1 > N)
    cout << -1 << endl;
  else {
    ll L = (N + 1) / 2;
    if (N == 2 * L - 1) {
      for (ll i = 0; i < L; ++i) {
        cout << K + 2 * i << " " << K + 3 * L - 2 - i << " "
             << K + 4 * L - 2 + i << endl;
      }
      for (ll i = 0; i < L - 1; ++i) {
        cout << K + 1 + 2 * i << " " << K + 4 * L - 3 - i << " "
             << K + 5 * L - 2 + i << endl;
      }
    } else {
      for (ll i = 0; i < L; ++i) {
        cout << K + 2 * i << " " << K + (i == 0 ? 2 : 3) * L - i << " "
             << K + 4 * L + i << endl;
      }
      for (ll i = 0; i < L; ++i) {
        cout << K + 1 + 2 * i << " " << K + 4 * L - 1 - i << " "
             << K + 5 * L + i << endl;
      }
    }
  }
  return 0;
}

```