#include <iostream>
#include <vector>
#include <algorithm>
#include <stack>
#include <set>
#include <queue>
#include <map>
using namespace std;
typedef long long int ll;

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	ll n,k; cin >> n >> k;
	vector<ll> a(n),b(n),c(n);
	for(ll i=0;i<n;i++){
		c[i]=k+2*n+i;
	}
	for(ll i=0;i<n;i+=2){
		a[i]=k+i/2;
		b[i]=k+n+(n+1)/2+i/2+(n%2^1)-1;
	}
	for(ll i=1;i<n;i+=2){
		a[i]=k+(n+1)/2+i/2;
		b[i]=k+n+i/2;
	}
	for(ll i=0;i<n;i++){
		if(c[i]<a[i]+b[i]){
			cout << -1 << endl;
			return 0;
		}
	}
	for(ll i=0;i<n;i++){
		cout << a[i] << " " << b[i] << " " << c[i] << endl;
	}
}
### Prompt

Create a solution in Cpp for 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 <iostream>
#include <vector>
#include <algorithm>
#include <stack>
#include <set>
#include <queue>
#include <map>
using namespace std;
typedef long long int ll;

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	ll n,k; cin >> n >> k;
	vector<ll> a(n),b(n),c(n);
	for(ll i=0;i<n;i++){
		c[i]=k+2*n+i;
	}
	for(ll i=0;i<n;i+=2){
		a[i]=k+i/2;
		b[i]=k+n+(n+1)/2+i/2+(n%2^1)-1;
	}
	for(ll i=1;i<n;i+=2){
		a[i]=k+(n+1)/2+i/2;
		b[i]=k+n+i/2;
	}
	for(ll i=0;i<n;i++){
		if(c[i]<a[i]+b[i]){
			cout << -1 << endl;
			return 0;
		}
	}
	for(ll i=0;i<n;i++){
		cout << a[i] << " " << b[i] << " " << c[i] << endl;
	}
}
```