#include <bits/stdc++.h>
using namespace std;
 
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	long long n, k; cin >> n >> k;
	if (k + k <= n + 1) {
		set<long long> s, ss;
		for (int i = n; i < n + n; i++) s.insert(k + i), ss.insert(k + i + n);
		int now = *s.rbegin();
		for (int i = n; i < n + k; i++) {
			cout << i << " " << now << " " << i + now << endl;
			s.erase(now);
			ss.erase(i + now);
			now -= 2;
		}
		for (int i = k; i < n; i++) {
			cout << i << " " << *s.rbegin() << " " << *ss.rbegin() << endl;
			ss.erase(*ss.rbegin());
			s.erase(*s.rbegin());
		}
	} else {
		cout << -1 << endl;
	}
	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 <bits/stdc++.h>
using namespace std;
 
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	long long n, k; cin >> n >> k;
	if (k + k <= n + 1) {
		set<long long> s, ss;
		for (int i = n; i < n + n; i++) s.insert(k + i), ss.insert(k + i + n);
		int now = *s.rbegin();
		for (int i = n; i < n + k; i++) {
			cout << i << " " << now << " " << i + now << endl;
			s.erase(now);
			ss.erase(i + now);
			now -= 2;
		}
		for (int i = k; i < n; i++) {
			cout << i << " " << *s.rbegin() << " " << *ss.rbegin() << endl;
			ss.erase(*ss.rbegin());
			s.erase(*s.rbegin());
		}
	} else {
		cout << -1 << endl;
	}
	return 0;
}
```