#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  long long n, k;
  cin >> n >> k;
  if (k * 2 - 1 > n) {
    cout << -1 << '\n';
    return 0;
  }
  vector<pair<long long, long long>> ans(n);
  for (int i = 0; i < n; i++) {
    if (i * 2 < n) {
      ans[i].first = k + i;
      ans[i].second = k + n + n / 2 + i;
    } else {
      ans[i].first = k + i;
      ans[i].second = k + n + i - (n + 1) / 2;
    }
    //cerr << ans[i].first << " " << ans[i].second << endl;
  }
  //cerr << "end\n";
  sort(ans.begin(), ans.end(), [](const pair<long long, long long>& x, const pair<long long, long long>& y) { return x.first + x.second < y.first + y.second;});
  for (int i = 0; i < n; i++) {
    //assert(ans[i].first + ans[i].second < k +  n * 2 - 1 + i);
    cout << ans[i].first << " " << ans[i].second << " " << k + n * 2 + i << '\n';
  }
  return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  long long n, k;
  cin >> n >> k;
  if (k * 2 - 1 > n) {
    cout << -1 << '\n';
    return 0;
  }
  vector<pair<long long, long long>> ans(n);
  for (int i = 0; i < n; i++) {
    if (i * 2 < n) {
      ans[i].first = k + i;
      ans[i].second = k + n + n / 2 + i;
    } else {
      ans[i].first = k + i;
      ans[i].second = k + n + i - (n + 1) / 2;
    }
    //cerr << ans[i].first << " " << ans[i].second << endl;
  }
  //cerr << "end\n";
  sort(ans.begin(), ans.end(), [](const pair<long long, long long>& x, const pair<long long, long long>& y) { return x.first + x.second < y.first + y.second;});
  for (int i = 0; i < n; i++) {
    //assert(ans[i].first + ans[i].second < k +  n * 2 - 1 + i);
    cout << ans[i].first << " " << ans[i].second << " " << k + n * 2 + i << '\n';
  }
  return 0;
}
```