#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef pair<int, int>pii;
int main()
{
	int n, k;
	scanf("%d%d", &n, &k);
	if (k + k > n + 1)
	{
		printf("-1\n");
		return 0;
	}
	vector<pii>v;
	for (int i = 0; i < n; i++)
	{
		int t;
		if (n % 2 == 0)
		{
			if (i % 2 == 0)t = k + n / 2 - 1 - i / 2;
			else t = k + n - 1 - i / 2;
		}
		else
		{
			if (i % 2 == 0)t = k + n - 1 - i / 2;
			else t = k + n / 2 - 1 - i / 2;
		}
		v.push_back(make_pair(k + n + i + t, t));
	}
	sort(v.begin(), v.end());
	for (int i = 0; i < n; i++)printf("%d %d %d\n", v[i].second, v[i].first - v[i].second, k + n + n + i);
}
### 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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef pair<int, int>pii;
int main()
{
	int n, k;
	scanf("%d%d", &n, &k);
	if (k + k > n + 1)
	{
		printf("-1\n");
		return 0;
	}
	vector<pii>v;
	for (int i = 0; i < n; i++)
	{
		int t;
		if (n % 2 == 0)
		{
			if (i % 2 == 0)t = k + n / 2 - 1 - i / 2;
			else t = k + n - 1 - i / 2;
		}
		else
		{
			if (i % 2 == 0)t = k + n - 1 - i / 2;
			else t = k + n / 2 - 1 - i / 2;
		}
		v.push_back(make_pair(k + n + i + t, t));
	}
	sort(v.begin(), v.end());
	for (int i = 0; i < n; i++)printf("%d %d %d\n", v[i].second, v[i].first - v[i].second, k + n + n + i);
}
```