#include<bits/stdc++.h>
using namespace std;
struct Ans{
    int a,b,c;
};
int main()
{
    int n,k;
    cin>>n>>k;
    if(n<2*k-1) puts("-1");
    else{
        vector<Ans> ans;
        int c=k+2*n,b=k+n+n/2-(n%2==0);
        for(int i=k;i<k+n;i+=2)
        {
            ans.push_back(Ans{i,b,c});
            c++;b--;
        }
        b=k+2*n-1;
        for(int i=k+1;i<k+n;i+=2)
        {
            ans.push_back(Ans{i,b,c});
            c++;
            b--;
        }
        for(auto u:ans) {
            cout<<u.a<<" "<<u.b<<" "<<u.c<<'\n';
        }

    }
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;
struct Ans{
    int a,b,c;
};
int main()
{
    int n,k;
    cin>>n>>k;
    if(n<2*k-1) puts("-1");
    else{
        vector<Ans> ans;
        int c=k+2*n,b=k+n+n/2-(n%2==0);
        for(int i=k;i<k+n;i+=2)
        {
            ans.push_back(Ans{i,b,c});
            c++;b--;
        }
        b=k+2*n-1;
        for(int i=k+1;i<k+n;i+=2)
        {
            ans.push_back(Ans{i,b,c});
            c++;
            b--;
        }
        for(auto u:ans) {
            cout<<u.a<<" "<<u.b<<" "<<u.c<<'\n';
        }

    }
}
```