#ifdef LOCAL_EXEC
#include "include.h"
#else
#include <bits/stdc++.h>
using namespace std;
#define trace(...)
#endif

using LL = long long;

int main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL);

	int n, k; cin >> n >> k;

	if (k > (n + 1) / 2) {
		cout << -1 << endl;
		return 0;
	}

	int a = k + n / 2, b = k + 2 * n - 1, c = k + 3 * n - 1;
	for (int i = 0; i < n; i++) {
		cout << a << ' ' << b << ' ' << c << '\n';
		if (a == k + n - 1) {
			a = k;
			b = k + 2 * n - 2;
		} else {
			a++;
			b -= 2;
		}
		c--;
	}

	return 0;
}

### Prompt

Generate 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
#ifdef LOCAL_EXEC
#include "include.h"
#else
#include <bits/stdc++.h>
using namespace std;
#define trace(...)
#endif

using LL = long long;

int main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL);

	int n, k; cin >> n >> k;

	if (k > (n + 1) / 2) {
		cout << -1 << endl;
		return 0;
	}

	int a = k + n / 2, b = k + 2 * n - 1, c = k + 3 * n - 1;
	for (int i = 0; i < n; i++) {
		cout << a << ' ' << b << ' ' << c << '\n';
		if (a == k + n - 1) {
			a = k;
			b = k + 2 * n - 2;
		} else {
			a++;
			b -= 2;
		}
		c--;
	}

	return 0;
}

```