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

int n, k;
int f(int ind, bool r){
	return k + r * n + ind;
}

int32_t main(){
	cin >> n >> k;
	if(k > (n + 1) / 2){
		cout << -1;
		return 0;
	}
	bool state = 0;
	int stL = n / 2;
	int stR = n - 1;
	for(int i = 0; i < n / 2 - (n % 2 == 0); i++){
		cout << f(stR, 1) << ' ' << f(stL, 0) << ' ' << (k + 3 * n - i * 2 - 1) << endl;
		stL--;
		cout << f(stR, 0) << ' ' << f(stL, 1) << ' ' << (k + 3 * n - i * 2 - 2) << endl;
		stR--;
	}
	cout << f(stR, 1) << ' ' << f(stL, 0) << ' ' << (k + 2 * n + (n % 2 == 0)) << endl;
	if(n % 2 == 0){
		cout << f(0, 0) << ' ' << f(0, 1) << ' ' << (k + 2 * n) << 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>
#define int long long
using namespace std;

int n, k;
int f(int ind, bool r){
	return k + r * n + ind;
}

int32_t main(){
	cin >> n >> k;
	if(k > (n + 1) / 2){
		cout << -1;
		return 0;
	}
	bool state = 0;
	int stL = n / 2;
	int stR = n - 1;
	for(int i = 0; i < n / 2 - (n % 2 == 0); i++){
		cout << f(stR, 1) << ' ' << f(stL, 0) << ' ' << (k + 3 * n - i * 2 - 1) << endl;
		stL--;
		cout << f(stR, 0) << ' ' << f(stL, 1) << ' ' << (k + 3 * n - i * 2 - 2) << endl;
		stR--;
	}
	cout << f(stR, 1) << ' ' << f(stL, 0) << ' ' << (k + 2 * n + (n % 2 == 0)) << endl;
	if(n % 2 == 0){
		cout << f(0, 0) << ' ' << f(0, 1) << ' ' << (k + 2 * n) << endl;
	}
}
```