#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
ll a[N], b[N], c[N];
int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	ll n, k;
	cin >> n >> k;
	for (int i = 0; i < n; i++) {
		c[i] = k + 3 * n - 1 - i;
		if (2 * i < n) {
			a[i] = k + n / 2 + i;
			b[i] = k + 2 * n - 1 - 2 * i;
		} else {
			int p = (2 * i - n) / 2;
			a[i] = k + p;
			b[i] = k + 2 * n - 2 - 2 * p;
		}
		if (a[i] + b[i] > c[i])
			return cout << "-1", 0;
	}
	for (int i = 0; i < n; i++) {
		cout << a[i] << " " << b[i] << " " << c[i] << "\n";
	}
	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;
typedef long long ll;
const int N = 1e5 + 5;
ll a[N], b[N], c[N];
int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	ll n, k;
	cin >> n >> k;
	for (int i = 0; i < n; i++) {
		c[i] = k + 3 * n - 1 - i;
		if (2 * i < n) {
			a[i] = k + n / 2 + i;
			b[i] = k + 2 * n - 1 - 2 * i;
		} else {
			int p = (2 * i - n) / 2;
			a[i] = k + p;
			b[i] = k + 2 * n - 2 - 2 * p;
		}
		if (a[i] + b[i] > c[i])
			return cout << "-1", 0;
	}
	for (int i = 0; i < n; i++) {
		cout << a[i] << " " << b[i] << " " << c[i] << "\n";
	}
	return 0;
}
```