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

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cerr.tie(nullptr);

	int n, k;
	cin >> n >> k;

	if (n < 2*k-1) {
		cout << "-1\n";
		return 0;
	}

	if (n % 2 == 1) {
		for (int i=0; i<n/2; i++) {
			cout << k+i << ' ' << k+2*n-n/2+i << ' ' << k+2*n+2*i+1 << '\n';
		}
		for (int i=n/2; i<n; i++) {
			cout << k+i << ' ' << k+n-n/2+i << ' ' << k+n+2*i+1 << '\n';
		}
	} else {
		for (int i=0; i<n/2; i++) {
			cout << k+i << ' ' << k+2*n-n/2+i << ' ' << k+2*n+2*i << '\n';
		}
		for (int i=n/2; i<n; i++) {
			cout << k+i << ' ' << k+n-n/2+i << ' ' << k+n+2*i+1 << '\n';
		}
	}
}
### Prompt

Your task is to create 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;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cerr.tie(nullptr);

	int n, k;
	cin >> n >> k;

	if (n < 2*k-1) {
		cout << "-1\n";
		return 0;
	}

	if (n % 2 == 1) {
		for (int i=0; i<n/2; i++) {
			cout << k+i << ' ' << k+2*n-n/2+i << ' ' << k+2*n+2*i+1 << '\n';
		}
		for (int i=n/2; i<n; i++) {
			cout << k+i << ' ' << k+n-n/2+i << ' ' << k+n+2*i+1 << '\n';
		}
	} else {
		for (int i=0; i<n/2; i++) {
			cout << k+i << ' ' << k+2*n-n/2+i << ' ' << k+2*n+2*i << '\n';
		}
		for (int i=n/2; i<n; i++) {
			cout << k+i << ' ' << k+n-n/2+i << ' ' << k+n+2*i+1 << '\n';
		}
	}
}
```