#include<bits/stdc++.h>
#include <array>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
using LL = long long;
using ULL = unsigned long long;


void solve() {
	unsigned n, k; cin >> n >> k;
	if (2 * k > n + 1) { cout << -1 << endl; return; }

	int d = (n + 1) / 2;
	for (int p = 0; p < d; p++) {
		int a = k + 2 * p;
		int b = k + n + d - p - 1;
		int c = k + p + 2 * n;
		cout << a << " " << b << " " << c << endl;
	}
	for (int p = d; p < n; p++) {
		int a = k + 2 * (p - d) + 1;
		int b = k + n + n - (p - d) - 1;
		int c = k + p + 2 * n;
		cout << a << " " << b << " " << c << endl;
	}
}

int main() {

	solve();

	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>
#include <array>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
using LL = long long;
using ULL = unsigned long long;


void solve() {
	unsigned n, k; cin >> n >> k;
	if (2 * k > n + 1) { cout << -1 << endl; return; }

	int d = (n + 1) / 2;
	for (int p = 0; p < d; p++) {
		int a = k + 2 * p;
		int b = k + n + d - p - 1;
		int c = k + p + 2 * n;
		cout << a << " " << b << " " << c << endl;
	}
	for (int p = d; p < n; p++) {
		int a = k + 2 * (p - d) + 1;
		int b = k + n + n - (p - d) - 1;
		int c = k + p + 2 * n;
		cout << a << " " << b << " " << c << endl;
	}
}

int main() {

	solve();

	return 0;
}

```