#include <iostream>
#include <algorithm>
#include <vector>
#include <math.h>
#include <tuple>
#include <time.h>
using namespace std;
typedef long long ll;

int main() {
    ll n, k;
    cin >> n >> k;
    if (n < 2 * k - 1) {
        cout << -1 << endl;
        return 0;
    }

    ll a[n + 5], b[n + 5], c[n + 5], t = (n + 1) / 2;
    ll q = 2 * n - 1, r = 3 * n - 1, p = r - t - q;
    for (int i = 0; i < n; i++) {
        a[i] = p;
        b[i] = q;
        c[i] = r;
        p++;
        q -= 2;
        r--;
        if (p >= q) {
            if (n % 2 == 1) {
                p = 0;
                q = r - t;
            }
            else {
                p = 0;
                q = r - t - 1;
            }
        }
    }

    for (int i = 0; i < n; i++) {
        cout << a[i] + k << " " << b[i] + k << " " << c[i] + k << endl;
    }
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <algorithm>
#include <vector>
#include <math.h>
#include <tuple>
#include <time.h>
using namespace std;
typedef long long ll;

int main() {
    ll n, k;
    cin >> n >> k;
    if (n < 2 * k - 1) {
        cout << -1 << endl;
        return 0;
    }

    ll a[n + 5], b[n + 5], c[n + 5], t = (n + 1) / 2;
    ll q = 2 * n - 1, r = 3 * n - 1, p = r - t - q;
    for (int i = 0; i < n; i++) {
        a[i] = p;
        b[i] = q;
        c[i] = r;
        p++;
        q -= 2;
        r--;
        if (p >= q) {
            if (n % 2 == 1) {
                p = 0;
                q = r - t;
            }
            else {
                p = 0;
                q = r - t - 1;
            }
        }
    }

    for (int i = 0; i < n; i++) {
        cout << a[i] + k << " " << b[i] + k << " " << c[i] + k << endl;
    }
}
```