#include <bits/stdc++.h>
using namespace std;
typedef vector<int> VI;
typedef pair <int,int> ii;
typedef long long LL;
#define pb push_back
const int INF = 2147483647;

int n, a1, a2, a3, a4, l1, l2, i, k;

int main() {
scanf("%d %d", &n, &k);
if (n < 2 * k - 1) {
	printf("-1\n");
	return 0;
}
a1 = k;
a2 = k + 2 * n - 1;
a3 = k + 2 * n;
a4 = k + 3 * n - 1;
l1 = n / 2 + n % 2;
for (i=0;i<l1;i++) printf("%d %d %d\n", a1 + i, a3 - l1 + i, a3 + i * 2);
l2 = n - l1;
for (i=0;i<l2;i++) printf("%d %d %d\n", a1 + l1 + i, a1 + l1 + i + l2, a3 + i * 2 + 1);
return 0;
}
### 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;
typedef vector<int> VI;
typedef pair <int,int> ii;
typedef long long LL;
#define pb push_back
const int INF = 2147483647;

int n, a1, a2, a3, a4, l1, l2, i, k;

int main() {
scanf("%d %d", &n, &k);
if (n < 2 * k - 1) {
	printf("-1\n");
	return 0;
}
a1 = k;
a2 = k + 2 * n - 1;
a3 = k + 2 * n;
a4 = k + 3 * n - 1;
l1 = n / 2 + n % 2;
for (i=0;i<l1;i++) printf("%d %d %d\n", a1 + i, a3 - l1 + i, a3 + i * 2);
l2 = n - l1;
for (i=0;i<l2;i++) printf("%d %d %d\n", a1 + l1 + i, a1 + l1 + i + l2, a3 + i * 2 + 1);
return 0;
}
```