#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
using ll = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, k;
    cin >> n >> k;

    int m = n / 2;
    int d[5] = { 0, m, 0, n - m, 0 };
    for (int i = 0; i < 5; i++) {
        d[i] += k + (i / 2) * n;
    }
    if (d[1] + d[2] > d[4]) {
        cout << -1 << endl;
        exit(0);
    }

    for (int i = 0; i < n - m; i++) {
        cout << d[1] + i << ' ' << d[2] + i << ' ' << d[4] + i * 2 << '\n';
    }
    for (int i = 0; i < m; i++) {
        cout << d[0] + i << ' ' << d[3] + i << ' ' << d[4] + i * 2 + 1 << '\n';
    }

    return 0;
}
### 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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
using ll = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, k;
    cin >> n >> k;

    int m = n / 2;
    int d[5] = { 0, m, 0, n - m, 0 };
    for (int i = 0; i < 5; i++) {
        d[i] += k + (i / 2) * n;
    }
    if (d[1] + d[2] > d[4]) {
        cout << -1 << endl;
        exit(0);
    }

    for (int i = 0; i < n - m; i++) {
        cout << d[1] + i << ' ' << d[2] + i << ' ' << d[4] + i * 2 << '\n';
    }
    for (int i = 0; i < m; i++) {
        cout << d[0] + i << ' ' << d[3] + i << ' ' << d[4] + i * 2 + 1 << '\n';
    }

    return 0;
}
```