#include<iostream>
#include<cstdio>
#include<vector>
#define ll long long
using namespace std;
ll n,k;
vector<ll>vec;
vector<ll>::iterator it;
int main(){
	scanf("%lld%lld",&n,&k);
	if((k+k+n+n-1)*(n+n)/2>(k+n+n+k+n+n+n-1)*n/2)return printf("-1"),0;
	int up=(n+1)>>1;
	for(register int i=0;i<up;i++)printf("%lld %lld %lld\n",k+i,k+n+n-up+i,k+n+n+(i<<1));
	for(register int i=up;i<n;i++)printf("%lld %lld %lld\n",k+i,k+n+i-up,k+n+n+((i-up)<<1)+1);
	return 0;
}
### Prompt

Generate 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<iostream>
#include<cstdio>
#include<vector>
#define ll long long
using namespace std;
ll n,k;
vector<ll>vec;
vector<ll>::iterator it;
int main(){
	scanf("%lld%lld",&n,&k);
	if((k+k+n+n-1)*(n+n)/2>(k+n+n+k+n+n+n-1)*n/2)return printf("-1"),0;
	int up=(n+1)>>1;
	for(register int i=0;i<up;i++)printf("%lld %lld %lld\n",k+i,k+n+n-up+i,k+n+n+(i<<1));
	for(register int i=up;i<n;i++)printf("%lld %lld %lld\n",k+i,k+n+i-up,k+n+n+((i-up)<<1)+1);
	return 0;
}
```