#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long n,k;
    cin>>n>>k;
    for(int i=1;i<=(n+1)/2;i++)
    {
        //cout<<2*k+n/2+2*n+1<<'\n';
        if(2*k+n/2+2*n-1>k+(3*n-1))
        {

            cout<<"-1"<<'\n';
            return 0;
        }
        cout<<k+2*n+1-2*i<<' '<<k+n/2+i-1<<' '<<k+3*n-i<<'\n';
    }
    for(int i=1;i<=n/2;i++)
        cout<<k+2*n-2*i<<' '<<i+k-1<<' '<<k+3*n-i-(n+1)/2<<'\n';
}

### 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<bits/stdc++.h>
using namespace std;
int main()
{
    long long n,k;
    cin>>n>>k;
    for(int i=1;i<=(n+1)/2;i++)
    {
        //cout<<2*k+n/2+2*n+1<<'\n';
        if(2*k+n/2+2*n-1>k+(3*n-1))
        {

            cout<<"-1"<<'\n';
            return 0;
        }
        cout<<k+2*n+1-2*i<<' '<<k+n/2+i-1<<' '<<k+3*n-i<<'\n';
    }
    for(int i=1;i<=n/2;i++)
        cout<<k+2*n-2*i<<' '<<i+k-1<<' '<<k+3*n-i-(n+1)/2<<'\n';
}

```