#include <bits/stdc++.h>

using namespace std;

int const N = 300006;

int main() {
	int n, K;
	cin >> n >> K;
	if (2 * K > n + 1) {
		puts("-1");
		return 0;
	}
	static bool vis[N];
	auto ans = vector<pair<pair<int, int>, int>>();
	int l = 0, r = n / 2 + n - (n % 2 == 0);
	if (l == r)
		++r;
	for (int i = K + 2 * n; i < K + 3 * n; ++i) {
		if (vis[l])
			++l;
		if (vis[r])
			++r;
		vis[l] = vis[r] = 1;
		ans.push_back({ { K + l, K + r }, i });
		l += 2, r--;
		if (l >= n)
			l -= n;
		if (r < n)
			r += n;
	}
	//cerr << "? " << ' ' << ans.size() << '\n';
	for (auto const& p : ans)
		cout << p.first.first << ' ' << p.first.second << ' ' << p.second << '\n';
}

### 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;

int const N = 300006;

int main() {
	int n, K;
	cin >> n >> K;
	if (2 * K > n + 1) {
		puts("-1");
		return 0;
	}
	static bool vis[N];
	auto ans = vector<pair<pair<int, int>, int>>();
	int l = 0, r = n / 2 + n - (n % 2 == 0);
	if (l == r)
		++r;
	for (int i = K + 2 * n; i < K + 3 * n; ++i) {
		if (vis[l])
			++l;
		if (vis[r])
			++r;
		vis[l] = vis[r] = 1;
		ans.push_back({ { K + l, K + r }, i });
		l += 2, r--;
		if (l >= n)
			l -= n;
		if (r < n)
			r += n;
	}
	//cerr << "? " << ' ' << ans.size() << '\n';
	for (auto const& p : ans)
		cout << p.first.first << ' ' << p.first.second << ' ' << p.second << '\n';
}

```