#include <bits/stdc++.h>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    if (2*k > n+1) {
        cout << -1 << endl;
        return 0;
    }
    int d = (n+1)/2 - k;
    for (int i = 0; i < (n+1)/2; ++i) {
        cout << (k+2*i) << " " << (2*n-i - d) << " " << (k + 2*n + i) << "\n";
    }
    for (int i = 0; i < (n-1)/2; ++i) {
        cout << (k + 2*i + 1) << " " << (2*n + (n-1)/2 - i - d) << " " << (k + 2*n + (n+1)/2 + i) << "\n";
    }
    if (n % 2 == 0) {
        cout << (k+n-1) << " " << (k+n) << " " << (k+3*n-1) << endl;
    }
    return 0;
}

### Prompt

Develop a solution in CPP 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 <bits/stdc++.h>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    if (2*k > n+1) {
        cout << -1 << endl;
        return 0;
    }
    int d = (n+1)/2 - k;
    for (int i = 0; i < (n+1)/2; ++i) {
        cout << (k+2*i) << " " << (2*n-i - d) << " " << (k + 2*n + i) << "\n";
    }
    for (int i = 0; i < (n-1)/2; ++i) {
        cout << (k + 2*i + 1) << " " << (2*n + (n-1)/2 - i - d) << " " << (k + 2*n + (n+1)/2 + i) << "\n";
    }
    if (n % 2 == 0) {
        cout << (k+n-1) << " " << (k+n) << " " << (k+3*n-1) << endl;
    }
    return 0;
}

```