#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, k;
    cin >> n >> k;
    int lim = 3 * n / 2;
    if (k > 2 * n - lim) {
        cout << -1 << endl;
        return 0;
    }
    int l = 0, r = (3 * n - 1) / 2, cnt = 2 * n;
    while (l < r) {
        cout << l + k << " " << r + k << " " << (cnt++) + k << "\n";
        l += 2;
        --r;
    }
    l = 1;
    r = 2 * n - 1;
    while (l < n) {
        cout << l + k << " " << r + k << " " << (cnt++) + k << "\n";
        l += 2;
        --r;
    }
    return 0;
}
### Prompt

In CPP, your task is to solve 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() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, k;
    cin >> n >> k;
    int lim = 3 * n / 2;
    if (k > 2 * n - lim) {
        cout << -1 << endl;
        return 0;
    }
    int l = 0, r = (3 * n - 1) / 2, cnt = 2 * n;
    while (l < r) {
        cout << l + k << " " << r + k << " " << (cnt++) + k << "\n";
        l += 2;
        --r;
    }
    l = 1;
    r = 2 * n - 1;
    while (l < n) {
        cout << l + k << " " << r + k << " " << (cnt++) + k << "\n";
        l += 2;
        --r;
    }
    return 0;
}
```