/* cerberus97 - Hanit Banga */

#include <iostream>
#include <iomanip>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 1e5 + 10;

int main() {
	fast_cin();
	int n, k;
	cin >> n >> k;
	int a = n / 2, b = 2 * n - 1, c = 3 * n - 1;
	vector<vector<int>> t(n);
	for (int i = 0; i < n; ++i) {
		if (a + k + b + k > c + k) {
			cout << -1 << endl;
			return 0;
		}
		t[i] = {a + k, b + k, c + k};
		--c;
		if (a + 1 < b - 2) {
			++a; b -= 2;
		} else {
			a = 0;
			b = 2 * n - 2;
		}
	}
	for (int i = 0; i < n; ++i) {
		for (auto &j : t[i]) {
			cout << j << ' ';
		}
		cout << '\n';
	}
}
### Prompt

Construct a cpp code solution to the problem outlined:

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
/* cerberus97 - Hanit Banga */

#include <iostream>
#include <iomanip>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 1e5 + 10;

int main() {
	fast_cin();
	int n, k;
	cin >> n >> k;
	int a = n / 2, b = 2 * n - 1, c = 3 * n - 1;
	vector<vector<int>> t(n);
	for (int i = 0; i < n; ++i) {
		if (a + k + b + k > c + k) {
			cout << -1 << endl;
			return 0;
		}
		t[i] = {a + k, b + k, c + k};
		--c;
		if (a + 1 < b - 2) {
			++a; b -= 2;
		} else {
			a = 0;
			b = 2 * n - 2;
		}
	}
	for (int i = 0; i < n; ++i) {
		for (auto &j : t[i]) {
			cout << j << ' ';
		}
		cout << '\n';
	}
}
```