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

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<tuple<int,int,int>> ans;
  for (int i = 0; i < n/2; i++) {
    ans.emplace_back(k+n/2-1-i, k+n+i*2+1, k+n*2+i);
  }
  for (int i = 0; i < (n+1)/2; i++) {
    ans.emplace_back(k+n-1-i, k+n+i*2, k+n*2+n/2+i);
  }
  for (auto t : ans) {
    int a, b, c;
    tie(a,b,c) = t;
    if (a+b > c) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (auto t : ans) {
    int a, b, c;
    tie(a,b,c) = t;
    cout << a << ' ' << b << ' ' << c << '\n';
  }
}

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

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<tuple<int,int,int>> ans;
  for (int i = 0; i < n/2; i++) {
    ans.emplace_back(k+n/2-1-i, k+n+i*2+1, k+n*2+i);
  }
  for (int i = 0; i < (n+1)/2; i++) {
    ans.emplace_back(k+n-1-i, k+n+i*2, k+n*2+n/2+i);
  }
  for (auto t : ans) {
    int a, b, c;
    tie(a,b,c) = t;
    if (a+b > c) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (auto t : ans) {
    int a, b, c;
    tie(a,b,c) = t;
    cout << a << ' ' << b << ' ' << c << '\n';
  }
}

```