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

int main() {

    int N, K;
    cin >> N >> K;
	if(2*K> N + 1){
		cout << -1;
		return 0;
	}
    
	int a, b, c, ai, bi;
	int c_count = 0;
	for(int i=0;i<N/2+N%2;i++){
		c = 2 * N + K + c_count;
		c_count++;
		a = K + i * 2;
		b = K + N + N / 2 + N % 2 - 1 - i;
		cout << a << " " << b << " " << c << endl;
	}
	for(int i=0;i<N/2;i++){
		c = 2 * N + K + c_count;
		c_count++;
		a = K + i * 2 + 1;
		b = K + N * 2 - 1 - i;
		cout << a << " " << b << " " << c << endl;
	}
}
### Prompt

Your task is to create 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 main() {

    int N, K;
    cin >> N >> K;
	if(2*K> N + 1){
		cout << -1;
		return 0;
	}
    
	int a, b, c, ai, bi;
	int c_count = 0;
	for(int i=0;i<N/2+N%2;i++){
		c = 2 * N + K + c_count;
		c_count++;
		a = K + i * 2;
		b = K + N + N / 2 + N % 2 - 1 - i;
		cout << a << " " << b << " " << c << endl;
	}
	for(int i=0;i<N/2;i++){
		c = 2 * N + K + c_count;
		c_count++;
		a = K + i * 2 + 1;
		b = K + N * 2 - 1 - i;
		cout << a << " " << b << " " << c << endl;
	}
}
```