#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

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

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

	vector<int> a(n),b(n),c(n);
	int num=k+3*n;
	rep(i,n) c[n-i-1]=--num;
	rep(i,n) b[(i+k)%n]=--num;

	vector<pair<int,int>> p(n);
	rep(i,n) p[i]={c[i]-b[i],i};
	sort(p.rbegin(),p.rend());

	rep(i,n) a[p[i].second]=--num;

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

	return 0;
}

### Prompt

Please formulate a cpp solution to 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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

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

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

	vector<int> a(n),b(n),c(n);
	int num=k+3*n;
	rep(i,n) c[n-i-1]=--num;
	rep(i,n) b[(i+k)%n]=--num;

	vector<pair<int,int>> p(n);
	rep(i,n) p[i]={c[i]-b[i],i};
	sort(p.rbegin(),p.rend());

	rep(i,n) a[p[i].second]=--num;

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

	return 0;
}

```