#include"bits/stdc++.h"
using namespace std;
typedef long long LL;
const int N=1e6;
const LL mod=998244353;
struct node{
	LL a,b,c,sum;
}a[N];
bool cmp(node x,node y){
	return x.sum<y.sum;
}
int main(){
	LL n,k;
	cin>>n>>k;
	if(n<2*k-1){
		cout<<-1<<endl;
		return 0;
	}
	for(int i=1;i<=k;i++){
		a[i].a=i+k-1;
		a[i].b=2*n-1+i;
		a[i].sum=a[i].a+a[i].b;
	}
	for(int i=k+1;i<=k+n-1;i++){
		a[i].a=i+k-1;
		a[i].b=i+(n-k)+k-1;
		a[i].sum=a[i].a+a[i].b;
	}
	sort(a+1,a+n+1,cmp);
	LL c=k+2*n;
	for(int i=1;i<=n;i++){
		//if(a[i].a+a[i].b>c+i-1)cout<<"XXX"<<endl;
		printf("%lld %lld %lld\n",a[i].a,a[i].b,c+i-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"bits/stdc++.h"
using namespace std;
typedef long long LL;
const int N=1e6;
const LL mod=998244353;
struct node{
	LL a,b,c,sum;
}a[N];
bool cmp(node x,node y){
	return x.sum<y.sum;
}
int main(){
	LL n,k;
	cin>>n>>k;
	if(n<2*k-1){
		cout<<-1<<endl;
		return 0;
	}
	for(int i=1;i<=k;i++){
		a[i].a=i+k-1;
		a[i].b=2*n-1+i;
		a[i].sum=a[i].a+a[i].b;
	}
	for(int i=k+1;i<=k+n-1;i++){
		a[i].a=i+k-1;
		a[i].b=i+(n-k)+k-1;
		a[i].sum=a[i].a+a[i].b;
	}
	sort(a+1,a+n+1,cmp);
	LL c=k+2*n;
	for(int i=1;i<=n;i++){
		//if(a[i].a+a[i].b>c+i-1)cout<<"XXX"<<endl;
		printf("%lld %lld %lld\n",a[i].a,a[i].b,c+i-1);
	}
}

```