#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)
#define dbg(S) { int f=1; for(auto x:S) cerr<<(f?"[":", ")<<x, f=0; cerr<<"]\n"; }

using namespace std;
using lint=long long;

int main(){
	lint n,k; cin>>n>>k;

	if(2*k>n+1){
		cout<<-1<<'\n';
		return 0;
	}

	vector<lint> a(n),b(n),c(n);
	rep(i,n){
		c[i]=k+2*n+i;
	}

	rep(i,k-1){
		b[n-k+i+1]=k+2*n-i-1;
	}
	rep(i,n-k+1){
		b[i]=k+2*n-k-i;
	}

	vector<pair<lint,int>> p(n);
	rep(i,n){
		p[i]={c[i]-b[i],i};
	}
	sort(p.begin(),p.end());
	rep(i,n){
		a[p[i].second]=k+i;
	}

	rep(i,n){
		cout<<a[i]<<' '<<b[i]<<' '<<c[i]<<'\n';
	}

	return 0;
}

### 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>

#define rep(i,n) for(int i=0;i<(n);i++)
#define dbg(S) { int f=1; for(auto x:S) cerr<<(f?"[":", ")<<x, f=0; cerr<<"]\n"; }

using namespace std;
using lint=long long;

int main(){
	lint n,k; cin>>n>>k;

	if(2*k>n+1){
		cout<<-1<<'\n';
		return 0;
	}

	vector<lint> a(n),b(n),c(n);
	rep(i,n){
		c[i]=k+2*n+i;
	}

	rep(i,k-1){
		b[n-k+i+1]=k+2*n-i-1;
	}
	rep(i,n-k+1){
		b[i]=k+2*n-k-i;
	}

	vector<pair<lint,int>> p(n);
	rep(i,n){
		p[i]={c[i]-b[i],i};
	}
	sort(p.begin(),p.end());
	rep(i,n){
		a[p[i].second]=k+i;
	}

	rep(i,n){
		cout<<a[i]<<' '<<b[i]<<' '<<c[i]<<'\n';
	}

	return 0;
}

```