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

int n, k;

int main () {
	scanf("%d%d", &n, &k);
	if (k > (n + 1 >> 1)) puts("-1");
	else {
		if (n & 1) {
			int m = n + 1 >> 1;
			for (int i = 0; i < m; i++) printf("%d %d %d\n", k + i, k + 3 * m - 2 + i, k + 4 * m - 2 + 2 * i);
			for (int i = 0; i < m - 1; i++) printf("%d %d %d\n", k + m + i, k + 2 * m - 1 + i, k + 4 * m - 1 + 2 * i);
		}
		else {
			int m = n >> 1;
			for (int i = 0; i < m; i++) printf("%d %d %d\n", k + i, k + 3 * m + i, k + 4 * m + 2 * i);
			for (int i = 0; i < m; i++) printf("%d %d %d\n", k + m + i, k + 2 * m + i, k + 4 * m + 1 + 2 * i);
		}
	}
	return 0;
}
### Prompt

Please formulate a cpp solution 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>
using namespace std;

int n, k;

int main () {
	scanf("%d%d", &n, &k);
	if (k > (n + 1 >> 1)) puts("-1");
	else {
		if (n & 1) {
			int m = n + 1 >> 1;
			for (int i = 0; i < m; i++) printf("%d %d %d\n", k + i, k + 3 * m - 2 + i, k + 4 * m - 2 + 2 * i);
			for (int i = 0; i < m - 1; i++) printf("%d %d %d\n", k + m + i, k + 2 * m - 1 + i, k + 4 * m - 1 + 2 * i);
		}
		else {
			int m = n >> 1;
			for (int i = 0; i < m; i++) printf("%d %d %d\n", k + i, k + 3 * m + i, k + 4 * m + 2 * i);
			for (int i = 0; i < m; i++) printf("%d %d %d\n", k + m + i, k + 2 * m + i, k + 4 * m + 1 + 2 * i);
		}
	}
	return 0;
}
```