//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
ll N,K;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N>>K;
	if(K<=(N+1)/2){
		if(N%2){
			REP(i,(N-1)/2){
				cout<<K+i<<' '<<K+(3*N+1)/2+i<<' '<<K+2*N+1+2*i<<endl;
			}
			REP(i,(N-1)/2+1){
				cout<<K+(N-1)/2+i<<' '<<K+N+i<<' '<<K+2*N+2*i<<endl;
			}
		}else{
			REP(i,N/2){
				cout<<K+i<<' '<<K+3*N/2+i<<' '<<K+2*N+1+2*i<<endl;
			}
			REP(i,N/2){
				cout<<K+N/2+i<<' '<<K+N+i<<' '<<K+2*N+2*i<<endl;
			}
		}
	}else{
		cout<<-1<<endl;
	}
}
### Prompt

Construct a CPP code solution to the problem outlined:

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
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
ll N,K;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N>>K;
	if(K<=(N+1)/2){
		if(N%2){
			REP(i,(N-1)/2){
				cout<<K+i<<' '<<K+(3*N+1)/2+i<<' '<<K+2*N+1+2*i<<endl;
			}
			REP(i,(N-1)/2+1){
				cout<<K+(N-1)/2+i<<' '<<K+N+i<<' '<<K+2*N+2*i<<endl;
			}
		}else{
			REP(i,N/2){
				cout<<K+i<<' '<<K+3*N/2+i<<' '<<K+2*N+1+2*i<<endl;
			}
			REP(i,N/2){
				cout<<K+N/2+i<<' '<<K+N+i<<' '<<K+2*N+2*i<<endl;
			}
		}
	}else{
		cout<<-1<<endl;
	}
}
```