#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i = 0;i<((ll)(n));i++)
#define reg(i,a,b) for(ll i = ((ll)(a));i<=((ll)(b));i++)
#define irep(i,n) for(ll i = ((ll)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(ll i = ((ll)(b));i>=((ll)(a));i--)
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

/*
*/

ll f(ll x){
	return x*(x+1)/2;
}

bool ok=true;
ll n,k;

int main(void){
	cin>>n>>k;
	if(f(k+2*n-1)-f(k-1)>f(k+3*n-1)-f(k+2*n-1))ok=false;
	// cerr<<f(k+2*n-1)-f(k-1)<<","<<f(k+3*n-1)-f(k+2*n-1)<<endl;
	if(!ok){
		cout<<-1<<endl;
		return 0;
	}
	if(n%2==0){
		rep(i,n/2)cout<<k+i<<" "<<k+2*n-2-i*2<<" "<<k+2*n+(n/2-i)-1<<endl;
		rep(i,n/2)cout<<k+i+n/2<<" "<<k+2*n-1-i*2<<" "<<k+2*n+(n-i)-1<<endl;
	}else{
		rep(i,n/2)cout<<k+i<<" "<<k+2*n-2-i*2<<" "<<k+2*n+(n/2-i)-1<<endl;
		rep(i,n/2+1)cout<<k+i+n/2<<" "<<k+2*n-1-i*2<<" "<<k+2*n+(n-i)-1<<endl;
	}
	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>
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i = 0;i<((ll)(n));i++)
#define reg(i,a,b) for(ll i = ((ll)(a));i<=((ll)(b));i++)
#define irep(i,n) for(ll i = ((ll)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(ll i = ((ll)(b));i>=((ll)(a));i--)
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

/*
*/

ll f(ll x){
	return x*(x+1)/2;
}

bool ok=true;
ll n,k;

int main(void){
	cin>>n>>k;
	if(f(k+2*n-1)-f(k-1)>f(k+3*n-1)-f(k+2*n-1))ok=false;
	// cerr<<f(k+2*n-1)-f(k-1)<<","<<f(k+3*n-1)-f(k+2*n-1)<<endl;
	if(!ok){
		cout<<-1<<endl;
		return 0;
	}
	if(n%2==0){
		rep(i,n/2)cout<<k+i<<" "<<k+2*n-2-i*2<<" "<<k+2*n+(n/2-i)-1<<endl;
		rep(i,n/2)cout<<k+i+n/2<<" "<<k+2*n-1-i*2<<" "<<k+2*n+(n-i)-1<<endl;
	}else{
		rep(i,n/2)cout<<k+i<<" "<<k+2*n-2-i*2<<" "<<k+2*n+(n/2-i)-1<<endl;
		rep(i,n/2+1)cout<<k+i+n/2<<" "<<k+2*n-1-i*2<<" "<<k+2*n+(n-i)-1<<endl;
	}
	return 0;
}
```