#include <bits/stdc++.h>

#define long long long int
using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;
    if (n < 2 * k - 1) {
        cout << -1;
        return 0;
    }
    vector<int> a(n);
    vector<int> b(n);
    vector<int> c(n);
    for (int i = 0; i < n; i++) {
        c[i] = k + 2 * n + i;
    }
    int n1 = (n + 1) / 2;
    int n2 = n - n1;
    for (int i = 0; i < n1; i++) {
        a[i] = k + 2 * i;
        b[i] = k + 2 * n - n2 - i -1 ;
    }
    for (int i = 0; i < n2; i++) {
        a[n1 + i] = k + 2 * i + 1;
        b[n1 + i] = k + 2 * n - i - 1;
    }
    for (int i = 0; i < n; i++) {
        cout << a[i] << " " << b[i] << " " << c[i] << "\n";
    }

    return 0;
}


// 2 3 4  5 6 7  8 9 10
// 8 2 6
// 9 4 5
// 10 3 7

// k = 3, n = 5
// 3 .. 17

// 13 3 10    4 6 11 12
// 14 5 9
// 15 7 8
// 16 4 12
// 17 6 11
### 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 <bits/stdc++.h>

#define long long long int
using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;
    if (n < 2 * k - 1) {
        cout << -1;
        return 0;
    }
    vector<int> a(n);
    vector<int> b(n);
    vector<int> c(n);
    for (int i = 0; i < n; i++) {
        c[i] = k + 2 * n + i;
    }
    int n1 = (n + 1) / 2;
    int n2 = n - n1;
    for (int i = 0; i < n1; i++) {
        a[i] = k + 2 * i;
        b[i] = k + 2 * n - n2 - i -1 ;
    }
    for (int i = 0; i < n2; i++) {
        a[n1 + i] = k + 2 * i + 1;
        b[n1 + i] = k + 2 * n - i - 1;
    }
    for (int i = 0; i < n; i++) {
        cout << a[i] << " " << b[i] << " " << c[i] << "\n";
    }

    return 0;
}


// 2 3 4  5 6 7  8 9 10
// 8 2 6
// 9 4 5
// 10 3 7

// k = 3, n = 5
// 3 .. 17

// 13 3 10    4 6 11 12
// 14 5 9
// 15 7 8
// 16 4 12
// 17 6 11
```