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

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

  int n, k;
  cin >> n >> k;
  long long sum = (2LL * k + n * 3 - 1) * 3 * n / 2;
  long long sumC = (2LL * k + n * 5 - 1) * n / 2;
  if (sumC * 2 < sum)
  {
    cout << -1 << endl;
    return 0;
  }

  int half = (n + 1) / 2;
  for (int i = 0; i < half; i++)
    cout << k + n - half + i << ' ' << k + n * 2 - 1 - i * 2 << ' ' << k + n * 3 - 1 - i << '\n';
  for (int i = 0; i < n - half; i++)
    cout << k + i << ' ' << k + n * 2 - 2 - i * 2 << ' ' << k + n * 3 - 1 - half - i << '\n';
}

### Prompt

In cpp, your task is to solve 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;

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

  int n, k;
  cin >> n >> k;
  long long sum = (2LL * k + n * 3 - 1) * 3 * n / 2;
  long long sumC = (2LL * k + n * 5 - 1) * n / 2;
  if (sumC * 2 < sum)
  {
    cout << -1 << endl;
    return 0;
  }

  int half = (n + 1) / 2;
  for (int i = 0; i < half; i++)
    cout << k + n - half + i << ' ' << k + n * 2 - 1 - i * 2 << ' ' << k + n * 3 - 1 - i << '\n';
  for (int i = 0; i < n - half; i++)
    cout << k + i << ' ' << k + n * 2 - 2 - i * 2 << ' ' << k + n * 3 - 1 - half - i << '\n';
}

```