#include <bits/stdc++.h>

using namespace std;

signed main() {
	int n, k; cin >> n >> k;
	if (n < 2 * k - 1) { // 構築不可
		cout << -1 << endl;
	}
	else { // 構築可能
		vector<int> a(n), b(n), c(n);
		for (int i = 0; i < n; ++i)
			c[i] = k + 2 * n + i;
		for (int i = 0; i < (n + 1) / 2; ++i)
			b[i] = k + 2 * i;
		for (int i = (n + 1) / 2; i < n; ++i)
			b[i] = k + 2 * (i - (n + 1) / 2) + 1;
		for (int i = 0; i < (n + 1) / 2; ++i)
			a[i] = k + n + (n - 1) / 2 - i;
		for (int i = (n + 1) / 2; i < n; ++i)
			a[i] = k + 2 * n - 1 - (i - (n + 1) / 2);
		for (int i = 0; i < n; ++i)
			cout << a[i] << " " << b[i] << " " << c[i] << endl;
	}
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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;

signed main() {
	int n, k; cin >> n >> k;
	if (n < 2 * k - 1) { // 構築不可
		cout << -1 << endl;
	}
	else { // 構築可能
		vector<int> a(n), b(n), c(n);
		for (int i = 0; i < n; ++i)
			c[i] = k + 2 * n + i;
		for (int i = 0; i < (n + 1) / 2; ++i)
			b[i] = k + 2 * i;
		for (int i = (n + 1) / 2; i < n; ++i)
			b[i] = k + 2 * (i - (n + 1) / 2) + 1;
		for (int i = 0; i < (n + 1) / 2; ++i)
			a[i] = k + n + (n - 1) / 2 - i;
		for (int i = (n + 1) / 2; i < n; ++i)
			a[i] = k + 2 * n - 1 - (i - (n + 1) / 2);
		for (int i = 0; i < n; ++i)
			cout << a[i] << " " << b[i] << " " << c[i] << endl;
	}
	return 0;
}
```