#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ri register int
const long double PI=3.141592653589793238462643383279;
const int N=500005,M=998244353;
int n,k,i,p;
int main()
{
	scanf("%d %d",&n,&k);
	if(n<2*k-1)
	{
		puts("-1");
		return 0;
	}
	for(i=k,p=k+2*n;i<k+(n+1)/2;++i,p+=2)
		printf("%d %d %d\n",i,i+n+n/2,p);
	for(i=k+(n+1)/2,p=k+2*n+1;i<k+n;++i,p+=2)
		printf("%d %d %d\n",i,i+n-n/2-(n&1),p);	
}

### Prompt

In CPP, your task is to solve 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ri register int
const long double PI=3.141592653589793238462643383279;
const int N=500005,M=998244353;
int n,k,i,p;
int main()
{
	scanf("%d %d",&n,&k);
	if(n<2*k-1)
	{
		puts("-1");
		return 0;
	}
	for(i=k,p=k+2*n;i<k+(n+1)/2;++i,p+=2)
		printf("%d %d %d\n",i,i+n+n/2,p);
	for(i=k+(n+1)/2,p=k+2*n+1;i<k+n;++i,p+=2)
		printf("%d %d %d\n",i,i+n-n/2-(n&1),p);	
}

```