#include <iostream>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
typedef long long ll;
set<int> s,t;
pair<pair<ll,ll>,ll> p[100010];
int main(){
	long long i,n,k;
	cin >> n >> k;
	if(k*n<=n*(n + 1)/2){
		if(n%2==0){
			for(i=0;i<n/2;i++){
				p[i] = {{i,2*n - 2 - 2*i},2*n - 1 - i};
			}
			for(i=n/2;i<n;i++){
				p[i] = {{i,3*n - 1 - 2*i},3*n - 1 - i}; 
			}
		}else{
			for(i=0;i<n/2;i++){
				p[i] = {{i,2*n - 2 - 2*i},2*n - 2 - i};
			}
			for(i=n/2;i<n;i++){
				p[i] = {{i,3*n - 2 - 2*i},3*n - 2 - i};
			}
		}
		for(i=0;i<n;i++){
			cout << p[i].first.first + k << " " << p[i].first.second + k << " " << p[i].second + k + (n + 1)/2 << endl;
		}
	}else{
		cout << -1 << endl;
	}
}
### 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 <iostream>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
typedef long long ll;
set<int> s,t;
pair<pair<ll,ll>,ll> p[100010];
int main(){
	long long i,n,k;
	cin >> n >> k;
	if(k*n<=n*(n + 1)/2){
		if(n%2==0){
			for(i=0;i<n/2;i++){
				p[i] = {{i,2*n - 2 - 2*i},2*n - 1 - i};
			}
			for(i=n/2;i<n;i++){
				p[i] = {{i,3*n - 1 - 2*i},3*n - 1 - i}; 
			}
		}else{
			for(i=0;i<n/2;i++){
				p[i] = {{i,2*n - 2 - 2*i},2*n - 2 - i};
			}
			for(i=n/2;i<n;i++){
				p[i] = {{i,3*n - 2 - 2*i},3*n - 2 - i};
			}
		}
		for(i=0;i<n;i++){
			cout << p[i].first.first + k << " " << p[i].first.second + k << " " << p[i].second + k + (n + 1)/2 << endl;
		}
	}else{
		cout << -1 << endl;
	}
}
```