#include <bits/stdc++.h>
using namespace std;

int main() {
    int n, k;
    cin >> n >> k;
    if (2 * k - 1 > n) printf("-1\n"), exit(0);
    int tmp = k;
    auto print = [&](int a, int b, int c) {
        printf("%d %d %d\n", tmp + a - 1, tmp + n + b - 1, tmp + n + n + c - 1);
    };
    k = (n + 1) >> 1;
    for (int i = 1; i <= k; i++) {
        print(i, n - k + i, i * 2 - 1);
    }
    for (int i = k + 1; i <= n; i++) {
        print(i, i - k, (i - k) * 2);
    }
    return 0;
}
/*
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
4 14
5 15
6 16
7 17
8 11
9 12
10 13
3 4 5 6 7
8 9 10 11 12
13 14 15 16 17
3 10

*/
### Prompt

Create a solution in Cpp for 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() {
    int n, k;
    cin >> n >> k;
    if (2 * k - 1 > n) printf("-1\n"), exit(0);
    int tmp = k;
    auto print = [&](int a, int b, int c) {
        printf("%d %d %d\n", tmp + a - 1, tmp + n + b - 1, tmp + n + n + c - 1);
    };
    k = (n + 1) >> 1;
    for (int i = 1; i <= k; i++) {
        print(i, n - k + i, i * 2 - 1);
    }
    for (int i = k + 1; i <= n; i++) {
        print(i, i - k, (i - k) * 2);
    }
    return 0;
}
/*
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
4 14
5 15
6 16
7 17
8 11
9 12
10 13
3 4 5 6 7
8 9 10 11 12
13 14 15 16 17
3 10

*/
```