#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)
 
int main(){
	int n, k;
	scanf("%d%d", &n, &k);
	if (n + 1 < k << 1){
		printf("-1\n");
		return 0;
	}
	for (int i = n - 1, j = n - 1, w = n >> 1; j >= 0; --i, j -= 2, ++w)printf("%d %d %d\n", k + w, k + n + j, k + 2 * n + i);
	for (int i = n / 2 - 1, j = n - 2, w = 0; j >= 0; --i, j -= 2, ++w)printf("%d %d %d\n", k + w, k + n + j, k + 2 * n + i);
}
### Prompt

In cpp, your task is to solve 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;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)
 
int main(){
	int n, k;
	scanf("%d%d", &n, &k);
	if (n + 1 < k << 1){
		printf("-1\n");
		return 0;
	}
	for (int i = n - 1, j = n - 1, w = n >> 1; j >= 0; --i, j -= 2, ++w)printf("%d %d %d\n", k + w, k + n + j, k + 2 * n + i);
	for (int i = n / 2 - 1, j = n - 2, w = 0; j >= 0; --i, j -= 2, ++w)printf("%d %d %d\n", k + w, k + n + j, k + 2 * n + i);
}
```