#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double D;
typedef pair<ll,ll> P;
#define M 1000000007
#define F first
#define S second
#define PB push_back
#define INF 100000000000000000
ll n,k;
int main(void){
    cin>>n>>k;
    if(2*k>n+1)cout<<-1<<endl;
    else{
		if(n%2){
			ll a=k,b=k+n+n/2LL,c=k+n*2LL;
			for(int i=0;i<n/2+1;i++){
				cout<<a<<' '<<b<<' '<<c<<endl;
				a+=2LL,b--,c++;
			}
			a=k+1,b=k+2LL*n-1LL;
			for(int i=0;i<n/2;i++){
				cout<<a<<' '<<b<<' '<<c<<endl;
				a+=2LL,b--,c++;
			}
		}else{
			ll a=k,b=k+n+n/2LL,c=k+n*2LL;
			for(int i=0;i<n/2;i++){
				cout<<a<<' '<<b<<' '<<c<<endl;
				a+=2LL,b--,c++;
			}
			cout<<k+n-1LL<<' '<<k+n<<' '<<c<<endl;
			c++;
			a=k+1,b=k+2LL*n-1LL;
			for(int i=0;i<n/2-1;i++){
				cout<<a<<' '<<b<<' '<<c<<endl;
				a+=2LL,b--,c++;
			}
		}
	}
    
}
### Prompt

Please provide a cpp coded solution 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>
using namespace std;
typedef long long ll;
typedef double D;
typedef pair<ll,ll> P;
#define M 1000000007
#define F first
#define S second
#define PB push_back
#define INF 100000000000000000
ll n,k;
int main(void){
    cin>>n>>k;
    if(2*k>n+1)cout<<-1<<endl;
    else{
		if(n%2){
			ll a=k,b=k+n+n/2LL,c=k+n*2LL;
			for(int i=0;i<n/2+1;i++){
				cout<<a<<' '<<b<<' '<<c<<endl;
				a+=2LL,b--,c++;
			}
			a=k+1,b=k+2LL*n-1LL;
			for(int i=0;i<n/2;i++){
				cout<<a<<' '<<b<<' '<<c<<endl;
				a+=2LL,b--,c++;
			}
		}else{
			ll a=k,b=k+n+n/2LL,c=k+n*2LL;
			for(int i=0;i<n/2;i++){
				cout<<a<<' '<<b<<' '<<c<<endl;
				a+=2LL,b--,c++;
			}
			cout<<k+n-1LL<<' '<<k+n<<' '<<c<<endl;
			c++;
			a=k+1,b=k+2LL*n-1LL;
			for(int i=0;i<n/2-1;i++){
				cout<<a<<' '<<b<<' '<<c<<endl;
				a+=2LL,b--,c++;
			}
		}
	}
    
}
```