#include <iostream>
#include <vector>
using namespace std;
using ll = long long;
ll N, K;

int main() {
  cin >> N >> K;
  if(N < K*2 - 1) {
    cout << -1;
    return 0;
  }
  vector<ll> a, b;
  if(N & 1) {
    for(int i = K; i < K + N; i += 2) a.push_back(i);
    for(int i = N/2; i >= 0; --i) b.push_back(i + K + N);
    for(int i = K+1; i < K + N; i += 2) a.push_back(i);
    for(int i = N-1; i > N / 2; --i) b.push_back(i + K + N);
  } else {
    for(int i = K+1; i < K + N; i += 2) a.push_back(i);
    for(int i = N/2-1; i >= 0; --i) b.push_back(i + K + N);
    for(int i = K; i < K + N; i += 2) a.push_back(i);
    for(int i = N-1; i >= N / 2; --i) b.push_back(i + K + N);
  }
  for(int i = 0; i < N; ++i) cout << a[i] << " " << b[i] << " " << i + K + N * 2 << '\n';
  return 0;
}
### Prompt

Please create a solution in cpp 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 <vector>
using namespace std;
using ll = long long;
ll N, K;

int main() {
  cin >> N >> K;
  if(N < K*2 - 1) {
    cout << -1;
    return 0;
  }
  vector<ll> a, b;
  if(N & 1) {
    for(int i = K; i < K + N; i += 2) a.push_back(i);
    for(int i = N/2; i >= 0; --i) b.push_back(i + K + N);
    for(int i = K+1; i < K + N; i += 2) a.push_back(i);
    for(int i = N-1; i > N / 2; --i) b.push_back(i + K + N);
  } else {
    for(int i = K+1; i < K + N; i += 2) a.push_back(i);
    for(int i = N/2-1; i >= 0; --i) b.push_back(i + K + N);
    for(int i = K; i < K + N; i += 2) a.push_back(i);
    for(int i = N-1; i >= N / 2; --i) b.push_back(i + K + N);
  }
  for(int i = 0; i < N; ++i) cout << a[i] << " " << b[i] << " " << i + K + N * 2 << '\n';
  return 0;
}
```