#include <iostream>
#include <set>
#include <tuple>
#include <vector>

int main() {
  int n, k;
  std::cin >> n >> k;
  std::set<int> s, t;
  for (int i = k; i < k + n; ++i) s.insert(i);
  std::vector<std::tuple<int, int, int>> ans;
  for (int z = k + n + n + n - 1, y = k + n + n - 1; z >= k + n + n; --z) {
    auto it = s.upper_bound(z - y);
    if (it == s.begin()) {
      std::cout << "-1\n";
      return 0;
    }
    int x = *prev(it);
    s.erase(x);
    ans.emplace_back(x, y, z);
    if ((y -= 2) < k + n) y = k + n + n - 2;
  }
  for (auto g : ans) {
    std::cout << std::get<0>(g) << ' ' << std::get<1>(g) << ' '
              << std::get<2>(g) << "\n";
  }
  return 0;
}

### Prompt

Please formulate a cpp solution 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 <iostream>
#include <set>
#include <tuple>
#include <vector>

int main() {
  int n, k;
  std::cin >> n >> k;
  std::set<int> s, t;
  for (int i = k; i < k + n; ++i) s.insert(i);
  std::vector<std::tuple<int, int, int>> ans;
  for (int z = k + n + n + n - 1, y = k + n + n - 1; z >= k + n + n; --z) {
    auto it = s.upper_bound(z - y);
    if (it == s.begin()) {
      std::cout << "-1\n";
      return 0;
    }
    int x = *prev(it);
    s.erase(x);
    ans.emplace_back(x, y, z);
    if ((y -= 2) < k + n) y = k + n + n - 2;
  }
  for (auto g : ans) {
    std::cout << std::get<0>(g) << ' ' << std::get<1>(g) << ' '
              << std::get<2>(g) << "\n";
  }
  return 0;
}

```