#include <bits/stdc++.h>

using namespace std;

int main() {
  int N, K;
  cin >> N >> K;

  vector<vector<int>> data(N);
  for (int i = 0; i < N; i++) data[i].resize(3);

  for (int i = 0; i < N; i++) {
    data[i][2] = K + 2 * N + i;
  }
  for (int i = 0; i < (N / 2); i++) {
    data[N / 2 - 1 - i][0] = K + i;
    data[N / 2 - 1 - i][1] = K + 2 * N - 2 - 2 * i;
  }

  int base = (N + 1) / 2;
  for (int i = 0; i < base; i++) {
    data[N - i - 1][0] = K + N / 2 + i;
    data[N - i - 1][1] = K + 2 * N - 1 - 2 * i;
  }

  bool ans = true;
  for (int i = 0; i < N; i++) {
    if (data[i][0] + data[i][1] > data[i][2]) ans = false;
  }
  if (!ans) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < N; i++) {
    cout << data[i][0] << " " << data[i][1] << " " << data[i][2] << endl;
  }

  return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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() {
  int N, K;
  cin >> N >> K;

  vector<vector<int>> data(N);
  for (int i = 0; i < N; i++) data[i].resize(3);

  for (int i = 0; i < N; i++) {
    data[i][2] = K + 2 * N + i;
  }
  for (int i = 0; i < (N / 2); i++) {
    data[N / 2 - 1 - i][0] = K + i;
    data[N / 2 - 1 - i][1] = K + 2 * N - 2 - 2 * i;
  }

  int base = (N + 1) / 2;
  for (int i = 0; i < base; i++) {
    data[N - i - 1][0] = K + N / 2 + i;
    data[N - i - 1][1] = K + 2 * N - 1 - 2 * i;
  }

  bool ans = true;
  for (int i = 0; i < N; i++) {
    if (data[i][0] + data[i][1] > data[i][2]) ans = false;
  }
  if (!ans) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < N; i++) {
    cout << data[i][0] << " " << data[i][1] << " " << data[i][2] << endl;
  }

  return 0;
}
```