#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int n, k;
int a[maxn], b[maxn], c[maxn];
set<int> s;
int main() {
	scanf("%d%d", &n, &k);
	if(2 * k > n + 1) return puts("-1"), 0;
	for(int i = 0; i < n; i++) a[i] = k + i;
	for(int i = 0; i < n; i++) b[i] = k + n + i;
	for(int i = 0; i < n; i++) s.insert(k + n + n + i);	
	for(int i = 0; i < n; i++) {
		int A = a[i], B = b[(i + k - 1) % n];
		int C = A + B;
		auto it = s.lower_bound(C);
		s.erase(it);
		printf("%d %d %d\n", A, B, *it);
	}
	return 0;
}


### Prompt

Your challenge is to write 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;
const int maxn = 1e5 + 10;
int n, k;
int a[maxn], b[maxn], c[maxn];
set<int> s;
int main() {
	scanf("%d%d", &n, &k);
	if(2 * k > n + 1) return puts("-1"), 0;
	for(int i = 0; i < n; i++) a[i] = k + i;
	for(int i = 0; i < n; i++) b[i] = k + n + i;
	for(int i = 0; i < n; i++) s.insert(k + n + n + i);	
	for(int i = 0; i < n; i++) {
		int A = a[i], B = b[(i + k - 1) % n];
		int C = A + B;
		auto it = s.lower_bound(C);
		s.erase(it);
		printf("%d %d %d\n", A, B, *it);
	}
	return 0;
}


```