#include <bits/stdc++.h>
#define MAX 100010
using namespace std;

vector<pair<int, int>> tri;

void getTri(int l, int r)
{
	if (l + 1 == r)
	{
		tri.push_back({l, r});
		return;
	}
	if (l + 3 == r)
	{
		tri.push_back({l, r - 1});
		tri.push_back({l + 1, r});
		return;
	}
	int n = ((r - l + 1) / 2 + 1) / 2;
	for (int i = 0; i < n; i++) tri.push_back({l + i, r - (n - 1) + i});
	int m = (l + r + 1) / 2;
	for (int i = m; i <= r - n; i++) tri.push_back({l + n + i - m, i});
}

int mycmp(const pair<int, int> a, const pair<int, int> b)
{
	return a.first + a.second < b.first + b.second;
}

int main(void)
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int N, K;
	cin >> N >> K;
	if (2 * K - 2 > N - 1)
	{
		cout << -1 << endl;
		return 0;
	}
	getTri(K, K + 2 * N - 1);
	sort(tri.begin(), tri.end(), mycmp);
	for (int i = 0; i < N; i++) cout << tri[i].first << " " << tri[i].second << " " << K + 2 * N + i << "\n";
	return 0;
}

### Prompt

In Cpp, your task is to solve 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>
#define MAX 100010
using namespace std;

vector<pair<int, int>> tri;

void getTri(int l, int r)
{
	if (l + 1 == r)
	{
		tri.push_back({l, r});
		return;
	}
	if (l + 3 == r)
	{
		tri.push_back({l, r - 1});
		tri.push_back({l + 1, r});
		return;
	}
	int n = ((r - l + 1) / 2 + 1) / 2;
	for (int i = 0; i < n; i++) tri.push_back({l + i, r - (n - 1) + i});
	int m = (l + r + 1) / 2;
	for (int i = m; i <= r - n; i++) tri.push_back({l + n + i - m, i});
}

int mycmp(const pair<int, int> a, const pair<int, int> b)
{
	return a.first + a.second < b.first + b.second;
}

int main(void)
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int N, K;
	cin >> N >> K;
	if (2 * K - 2 > N - 1)
	{
		cout << -1 << endl;
		return 0;
	}
	getTri(K, K + 2 * N - 1);
	sort(tri.begin(), tri.end(), mycmp);
	for (int i = 0; i < N; i++) cout << tri[i].first << " " << tri[i].second << " " << K + 2 * N + i << "\n";
	return 0;
}

```