#include <bits/stdc++.h>

using namespace std;

int main(){
	ios_base::sync_with_stdio(false);
	long long n, k;
	cin >> n >> k;
	long long a = 2*n*k + (2*n) * (2 * n - 1) / 2;
	long long b = n * (k + 2*n) + (n) * (n - 1) / 2;
	
	if(a > b){
		cout << -1;
		return 0;
	}
	
	long long h = (n / 2);
	long long c = 0;
	typedef pair<long long, long long> ii;
	ii arr[n];
	for(int i = 0;i < n;i++){
		long long x = k + i;
		long long y = k + n + ((i + h) % n);
		arr[i] = ii(x+y,x);
	}
	
	sort(arr,arr+n);
	for(int i = 0;i < n;i++){
		cout << arr[i].second << " " << arr[i].first  - arr[i].second << " " << k + 2*n + i << "\n";
	}
	
}

### Prompt

Generate 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(){
	ios_base::sync_with_stdio(false);
	long long n, k;
	cin >> n >> k;
	long long a = 2*n*k + (2*n) * (2 * n - 1) / 2;
	long long b = n * (k + 2*n) + (n) * (n - 1) / 2;
	
	if(a > b){
		cout << -1;
		return 0;
	}
	
	long long h = (n / 2);
	long long c = 0;
	typedef pair<long long, long long> ii;
	ii arr[n];
	for(int i = 0;i < n;i++){
		long long x = k + i;
		long long y = k + n + ((i + h) % n);
		arr[i] = ii(x+y,x);
	}
	
	sort(arr,arr+n);
	for(int i = 0;i < n;i++){
		cout << arr[i].second << " " << arr[i].first  - arr[i].second << " " << k + 2*n + i << "\n";
	}
	
}

```