#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,k;
void work()
{
	if(n&1)
	{
		int k1=(n+1)/2;
		int i,j;
		for(i=k1,j=2*n;j<k1+2*n;++i,++j)printf("%d %d %d\n",i-k1+k,j-k1+k,i+j-k1+k);
		for(j=2*n+k1+1-i;i+j<k1+3*n;++i,++j)printf("%d %d %d\n",i-k1+k,j-k1+k,i+j-k1+k);
	}
	else
	{
		int k1=(n+1)/2;
		int i,j;
		for(i=k1,j=2*n;j<k1+2*n;++i,++j)printf("%d %d %d\n",i-k1+k,j-k1+k,i+j-k1+k);
		for(j=2*n+k1+1-i;i+j<k1+3*n-2;++i,++j)printf("%d %d %d\n",i-k1+k,j-k1+k,i+j-k1+k);
		printf("%d %d %d\n",i-k1+k,i+1-k1+k,i+j-k1+k);
	}
}
int main()
{
	scanf("%d %d",&n,&k);
	if(2*k<=n+1)work();
	else printf("-1");
	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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,k;
void work()
{
	if(n&1)
	{
		int k1=(n+1)/2;
		int i,j;
		for(i=k1,j=2*n;j<k1+2*n;++i,++j)printf("%d %d %d\n",i-k1+k,j-k1+k,i+j-k1+k);
		for(j=2*n+k1+1-i;i+j<k1+3*n;++i,++j)printf("%d %d %d\n",i-k1+k,j-k1+k,i+j-k1+k);
	}
	else
	{
		int k1=(n+1)/2;
		int i,j;
		for(i=k1,j=2*n;j<k1+2*n;++i,++j)printf("%d %d %d\n",i-k1+k,j-k1+k,i+j-k1+k);
		for(j=2*n+k1+1-i;i+j<k1+3*n-2;++i,++j)printf("%d %d %d\n",i-k1+k,j-k1+k,i+j-k1+k);
		printf("%d %d %d\n",i-k1+k,i+1-k1+k,i+j-k1+k);
	}
}
int main()
{
	scanf("%d %d",&n,&k);
	if(2*k<=n+1)work();
	else printf("-1");
	return 0;
}

```