#include <bits/stdc++.h>
using namespace std;
struct node{
	int x,y;
	int z;
};
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	int n,k;
	cin>>n>>k;
	int ave=k+k+n+n-1;
	vector<node>a(n);
	for(int i=0;i<n;i++){
		a[i].y=k+n+n-1-i;
	}
	int cha=n/2;
	for(int i=0;i<=cha;i++){
		a[i].x=k+cha-i;
	}
	int tmp=n-cha-2;
	for(int i=cha+1;i<n;i++){
		a[i].x=k+cha+1+(tmp-(i-cha-1));
	}
	sort(a.begin(),a.end(),[&](node a,node b){return a.x+a.y<b.x+b.y;});
	for(int i=0;i<n;i++)
		a[i].z=k+n+n+i;
	for(int i=0;i<n;i++){
		if(a[i].z<a[i].x+a[i].y){cout<<"-1\n";exit(0);}
	}
	for(int i=0;i<n;i++)
		cout<<a[i].x<<' '<<a[i].y<<' '<<a[i].z<<endl;
	return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;
struct node{
	int x,y;
	int z;
};
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	int n,k;
	cin>>n>>k;
	int ave=k+k+n+n-1;
	vector<node>a(n);
	for(int i=0;i<n;i++){
		a[i].y=k+n+n-1-i;
	}
	int cha=n/2;
	for(int i=0;i<=cha;i++){
		a[i].x=k+cha-i;
	}
	int tmp=n-cha-2;
	for(int i=cha+1;i<n;i++){
		a[i].x=k+cha+1+(tmp-(i-cha-1));
	}
	sort(a.begin(),a.end(),[&](node a,node b){return a.x+a.y<b.x+b.y;});
	for(int i=0;i<n;i++)
		a[i].z=k+n+n+i;
	for(int i=0;i<n;i++){
		if(a[i].z<a[i].x+a[i].y){cout<<"-1\n";exit(0);}
	}
	for(int i=0;i<n;i++)
		cout<<a[i].x<<' '<<a[i].y<<' '<<a[i].z<<endl;
	return 0;
}
```