#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(){
    int N, K;
    while(cin >> N >> K){
        vector<pair<int,int>> a(N);
        int m = 2*N;
        m = max(K+N, m);
        m = min(K+2*N-1, m);
        for(int i=0;i<N;i++){
            a[i] = make_pair(K+i, m);
            m++;
            if(m >= K+2*N) m -= N;
        }
        sort(a.begin(), a.end(), [&](pair<int,int> l, pair<int,int> r){return l.first+l.second < r.first+r.second;});
        bool ok = true;
        for(int i=0;i<N;i++){
            if(a[i].first+a[i].second > K+2*N+i) ok = false;
        }
        if(ok){
            for(int i=0;i<N;i++){
                cout << a[i].first << " " << a[i].second << " " << K+2*N+i << endl;
            }
        } else {
            cout << -1 << endl;
        }
    }
}
### Prompt

Please create a solution in CPP to 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>

using namespace std;

int main(){
    int N, K;
    while(cin >> N >> K){
        vector<pair<int,int>> a(N);
        int m = 2*N;
        m = max(K+N, m);
        m = min(K+2*N-1, m);
        for(int i=0;i<N;i++){
            a[i] = make_pair(K+i, m);
            m++;
            if(m >= K+2*N) m -= N;
        }
        sort(a.begin(), a.end(), [&](pair<int,int> l, pair<int,int> r){return l.first+l.second < r.first+r.second;});
        bool ok = true;
        for(int i=0;i<N;i++){
            if(a[i].first+a[i].second > K+2*N+i) ok = false;
        }
        if(ok){
            for(int i=0;i<N;i++){
                cout << a[i].first << " " << a[i].second << " " << K+2*N+i << endl;
            }
        } else {
            cout << -1 << endl;
        }
    }
}
```