#include <cstdio>
int n,k,i;
int main(){
	scanf("%d%d",&n,&k);
	if(2*k>n+1)puts("-1");
	else for(;i<n;i++)printf("%d %d %d\n", k+i,k+(i<(n+1)/2)*n+n/2+i,k+2*n+i*2-(i>=(n+1)/2)*((n+1)/2*2-1));
}
### 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 <cstdio>
int n,k,i;
int main(){
	scanf("%d%d",&n,&k);
	if(2*k>n+1)puts("-1");
	else for(;i<n;i++)printf("%d %d %d\n", k+i,k+(i<(n+1)/2)*n+n/2+i,k+2*n+i*2-(i>=(n+1)/2)*((n+1)/2*2-1));
}
```