#include <iostream>
#include <vector>
#include <algorithm>
#include <cassert>
using namespace std;
using ll = long long;
using Pa = pair<ll,ll>;
template <typename T> using vec = vector<T>;
template <typename T> using vvec = vector<vec<T>>;

int main(){
    ll N,K;
    cin >> N >> K;
    if(N+1<2*K){
        cout << -1 << endl;
        return 0;
    }
    vec<Pa> ab;
    for(int i=0;i<N/2;i++){
        ab.emplace_back(K+i,K+N+(N+1)/2+i);
    }
    for(int i=N/2;i<N;i++){
        ab.emplace_back(K+i,K+N+(i-N/2));
    }
    sort(ab.begin(),ab.end(),[](Pa& a, Pa& b){
        return a.first+a.second<b.first+b.second;
    });
    for(int i=0;i<N;i++){
        cout << ab[i].first << " " << ab[i].second << " " << K+2*N+i << endl;
        assert(ab[i].first+ab[i].second<=K+2*N+i);
    }
}
### 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 <cassert>
using namespace std;
using ll = long long;
using Pa = pair<ll,ll>;
template <typename T> using vec = vector<T>;
template <typename T> using vvec = vector<vec<T>>;

int main(){
    ll N,K;
    cin >> N >> K;
    if(N+1<2*K){
        cout << -1 << endl;
        return 0;
    }
    vec<Pa> ab;
    for(int i=0;i<N/2;i++){
        ab.emplace_back(K+i,K+N+(N+1)/2+i);
    }
    for(int i=N/2;i<N;i++){
        ab.emplace_back(K+i,K+N+(i-N/2));
    }
    sort(ab.begin(),ab.end(),[](Pa& a, Pa& b){
        return a.first+a.second<b.first+b.second;
    });
    for(int i=0;i<N;i++){
        cout << ab[i].first << " " << ab[i].second << " " << K+2*N+i << endl;
        assert(ab[i].first+ab[i].second<=K+2*N+i);
    }
}
```