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

#define fi first
#define se second
#define long long long
typedef pair<int,int> ii;

int A[100003], B[100003], C[100003];
int X[100003], Y[100003], Z[100003];

int main()
{
	// freopen("input.in", "r", stdin);

	int n, k; scanf("%d %d", &n, &k);
	for(int i = 1; i <= n; i++)
	{
		A[i] = k+i-1;
		B[i] = k+n+i-1;
		C[i] = k+2*n+i-1;
	}

	if(n&1 || 1)
	{
		int mid = (n+1)/2;
		for(int i = 1; i <= mid; i++)
			X[i] = A[i], Y[i] = B[i+mid-(n&1)], Z[i] = C[i*2-1];
		for(int i = mid+1; i <= n; i++)
			X[i] = A[i], Y[i] = B[i-mid], Z[i] = C[(i-mid)*2];

		for(int i = 1; i <= n; i++)
		{
			if(Z[i] < X[i]+Y[i])
			{
				printf("-1\n");
				return 0;
			}
		}
		for(int i = 1; i <= n; i++)
			printf("%d %d %d\n", X[i], Y[i], Z[i]);

	}
}
### 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;

#define fi first
#define se second
#define long long long
typedef pair<int,int> ii;

int A[100003], B[100003], C[100003];
int X[100003], Y[100003], Z[100003];

int main()
{
	// freopen("input.in", "r", stdin);

	int n, k; scanf("%d %d", &n, &k);
	for(int i = 1; i <= n; i++)
	{
		A[i] = k+i-1;
		B[i] = k+n+i-1;
		C[i] = k+2*n+i-1;
	}

	if(n&1 || 1)
	{
		int mid = (n+1)/2;
		for(int i = 1; i <= mid; i++)
			X[i] = A[i], Y[i] = B[i+mid-(n&1)], Z[i] = C[i*2-1];
		for(int i = mid+1; i <= n; i++)
			X[i] = A[i], Y[i] = B[i-mid], Z[i] = C[(i-mid)*2];

		for(int i = 1; i <= n; i++)
		{
			if(Z[i] < X[i]+Y[i])
			{
				printf("-1\n");
				return 0;
			}
		}
		for(int i = 1; i <= n; i++)
			printf("%d %d %d\n", X[i], Y[i], Z[i]);

	}
}
```