#include <bits/stdc++.h>
using namespace std;
#define all(V) V.begin(),V.end()

int main(){
	int N,K;
	cin>>N>>K;
	if(K>(N+1)/2){
		cout<<-1<<endl;
		return 0;
	}
	vector<tuple<int,int,int>> V(N);
	int S=(N+1)/2,L=S+N-1,M=2*N;
	for(int i=0;i<N;i++){
		get<1>(V[i])=L;
		get<2>(V[i])=M;
		get<0>(V[i])=L+M;
		
		L--;
		M--;
		if(M<S+N)M=S+2*N-1;
	}
	sort(all(V));
	for(int i=0;i<N;i++){
		get<0>(V[i])=K+2*N+i;
		get<1>(V[i])-=S-K;
		get<2>(V[i])-=S-K;
	}
	
	for(auto x:V){
		cout<<get<2>(x)<<" "<<get<1>(x)<<" "<<get<0>(x)<<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
#include <bits/stdc++.h>
using namespace std;
#define all(V) V.begin(),V.end()

int main(){
	int N,K;
	cin>>N>>K;
	if(K>(N+1)/2){
		cout<<-1<<endl;
		return 0;
	}
	vector<tuple<int,int,int>> V(N);
	int S=(N+1)/2,L=S+N-1,M=2*N;
	for(int i=0;i<N;i++){
		get<1>(V[i])=L;
		get<2>(V[i])=M;
		get<0>(V[i])=L+M;
		
		L--;
		M--;
		if(M<S+N)M=S+2*N-1;
	}
	sort(all(V));
	for(int i=0;i<N;i++){
		get<0>(V[i])=K+2*N+i;
		get<1>(V[i])-=S-K;
		get<2>(V[i])-=S-K;
	}
	
	for(auto x:V){
		cout<<get<2>(x)<<" "<<get<1>(x)<<" "<<get<0>(x)<<endl;
	}
}

```