#include<bits/stdc++.h>
using namespace std;
const int N=100005;
int n,k,a[N],b[N],c[N],tot;
int main(){
	scanf("%d%d",&n,&k);k--;
	for (int i=1;i<=n/2;i++){
		a[++tot]=i;
		b[tot]=i+n+n/2;
		a[++tot]=i+(n+1)/2;
		b[tot]=i+n;
	}
	if (n&1)a[++tot]=(n+1)/2,b[tot]=n+n;
	for (int i=1;i<=n;i++)c[i]=2*n+k+i,a[i]+=k,b[i]+=k;
	for (int i=1;i<=n;i++)
		if ((long long)a[i]+b[i]>c[i]){
			puts("-1");
			return 0;
		}
	for (int i=1;i<=n;i++)printf("%d %d %d\n",a[i],b[i],c[i]);	
}
### Prompt

Please provide a cpp coded solution 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<bits/stdc++.h>
using namespace std;
const int N=100005;
int n,k,a[N],b[N],c[N],tot;
int main(){
	scanf("%d%d",&n,&k);k--;
	for (int i=1;i<=n/2;i++){
		a[++tot]=i;
		b[tot]=i+n+n/2;
		a[++tot]=i+(n+1)/2;
		b[tot]=i+n;
	}
	if (n&1)a[++tot]=(n+1)/2,b[tot]=n+n;
	for (int i=1;i<=n;i++)c[i]=2*n+k+i,a[i]+=k,b[i]+=k;
	for (int i=1;i<=n;i++)
		if ((long long)a[i]+b[i]>c[i]){
			puts("-1");
			return 0;
		}
	for (int i=1;i<=n;i++)printf("%d %d %d\n",a[i],b[i],c[i]);	
}
```