#include "bits/stdc++.h"
using namespace std;

int main() {
	long long N, K;
	cin >> N >> K;
	if (2 * K > N + 1) {
		cout << -1 << endl;
		return 0;
	}
	vector<pair<long long, pair<long long, long long> > > ANS;
	for (int i = K; i <= N; i++) {
		ANS.push_back({ K + N + i * 2 - 1, { i, K + N + i - 1 } });
	}
	for (int i = N + 1; i < N + K; i++) {
		ANS.push_back({ i * 2 + K - 1 , { i, i + K - 1 } });
	}
	sort(ANS.begin(), ANS.end());
	for (int i = 0; i < N; i++) {
		cout << ANS[i].second.first << " " << ANS[i].second.second << " " << N * 2 + K + i << endl;
	}
}
### Prompt

Create a solution in CPP for 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 "bits/stdc++.h"
using namespace std;

int main() {
	long long N, K;
	cin >> N >> K;
	if (2 * K > N + 1) {
		cout << -1 << endl;
		return 0;
	}
	vector<pair<long long, pair<long long, long long> > > ANS;
	for (int i = K; i <= N; i++) {
		ANS.push_back({ K + N + i * 2 - 1, { i, K + N + i - 1 } });
	}
	for (int i = N + 1; i < N + K; i++) {
		ANS.push_back({ i * 2 + K - 1 , { i, i + K - 1 } });
	}
	sort(ANS.begin(), ANS.end());
	for (int i = 0; i < N; i++) {
		cout << ANS[i].second.first << " " << ANS[i].second.second << " " << N * 2 + K + i << endl;
	}
}
```