#include <bits/stdc++.h>
 
using namespace std;
typedef long long ll;
const int MAXN = 100100;

int K, N;

pair <int, int> pp[MAXN];

int main()
{
    ios_base::sync_with_stdio(0);
    cin >> K >> N;
    if (N > (K + 1) / 2)
    {
        cout << "-1\n";
        return 0;
    }

    int lo = K / 2;
    for (int i = 0; i < lo; i++)
    {
        int nv = (i + K - lo) + (N + K);
        pp[i] = make_pair (nv + i + N, i + N);
    }
    for (int i = lo; i < K; i++)
    {
        int nv = (i - lo) + (N + K);
        pp[i] = make_pair (nv + i + N, i + N);
    }

    sort (pp, pp + K);
    for (int i = 0; i < K; i++)
    {
        cout << pp[i].second << " " << pp[i].first - pp[i].second << " " << (N + 2 * K + i) << "\n";
    }
}
### Prompt

Please formulate a cpp solution 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;
typedef long long ll;
const int MAXN = 100100;

int K, N;

pair <int, int> pp[MAXN];

int main()
{
    ios_base::sync_with_stdio(0);
    cin >> K >> N;
    if (N > (K + 1) / 2)
    {
        cout << "-1\n";
        return 0;
    }

    int lo = K / 2;
    for (int i = 0; i < lo; i++)
    {
        int nv = (i + K - lo) + (N + K);
        pp[i] = make_pair (nv + i + N, i + N);
    }
    for (int i = lo; i < K; i++)
    {
        int nv = (i - lo) + (N + K);
        pp[i] = make_pair (nv + i + N, i + N);
    }

    sort (pp, pp + K);
    for (int i = 0; i < K; i++)
    {
        cout << pp[i].second << " " << pp[i].first - pp[i].second << " " << (N + 2 * K + i) << "\n";
    }
}
```