#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

long long N, K;
pair<long long, long long> A[1 << 18];

int main() {
	cin >> N >> K; long long cnt = K - 1;
	for (int i = 0; i < N; i++) { cnt++; A[i].first += cnt; }
	for (int i = N / 2; i < N; i++) { cnt++; A[i].first += cnt; A[i].second = cnt; }
	for (int i = 0; i < N / 2; i++) { cnt++; A[i].first += cnt; A[i].second = cnt; }
	sort(A, A + N);

	long long ret = cnt;
	for (int i = 0; i < N; i++) {
		ret++;
		if (A[i].first > ret) { cout << "-1" << endl; return 0; }
	}
	for (int i = 0; i < N; i++) {
		cnt++;
		cout << A[i].first - A[i].second << " " << A[i].second << " " << cnt << endl;
	}
	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
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

long long N, K;
pair<long long, long long> A[1 << 18];

int main() {
	cin >> N >> K; long long cnt = K - 1;
	for (int i = 0; i < N; i++) { cnt++; A[i].first += cnt; }
	for (int i = N / 2; i < N; i++) { cnt++; A[i].first += cnt; A[i].second = cnt; }
	for (int i = 0; i < N / 2; i++) { cnt++; A[i].first += cnt; A[i].second = cnt; }
	sort(A, A + N);

	long long ret = cnt;
	for (int i = 0; i < N; i++) {
		ret++;
		if (A[i].first > ret) { cout << "-1" << endl; return 0; }
	}
	for (int i = 0; i < N; i++) {
		cnt++;
		cout << A[i].first - A[i].second << " " << A[i].second << " " << cnt << endl;
	}
	return 0;
}
```