#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int vis[3 * N];
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	int n, K;
	cin >> n >> K;
	if(2 * K <= n + 1){
		for(int i = 2 * n; i < K + 2 * n; i++){
			cout << K + i - 2 * n << ' ' << i << ' ' << K + 2 * i - 2 * n << "\n";
			vis[i - 2 * n] = 1;
			vis[i - K] = 1;
			vis[2 * i - 2 * n] = 1;
		}
		vector<int> a, b, c;
		for(int i = 0; i < n; i++) if(!vis[i])
			a.push_back(K + i);
		for(int i = n; i < 2 * n; i++) if(!vis[i])
			b.push_back(K + i);
		for(int i = 2 * n; i < 3 * n; i++) if(!vis[i])
			c.push_back(K + i);
		for(int i = a.size() - 1; i >= 0; i--)
			cout << a[i] << ' ' << b[i] << ' ' << c[i] << "\n";
	}
	else{
		cout << "-1\n";
	}
	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 N = 100010;
int vis[3 * N];
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	int n, K;
	cin >> n >> K;
	if(2 * K <= n + 1){
		for(int i = 2 * n; i < K + 2 * n; i++){
			cout << K + i - 2 * n << ' ' << i << ' ' << K + 2 * i - 2 * n << "\n";
			vis[i - 2 * n] = 1;
			vis[i - K] = 1;
			vis[2 * i - 2 * n] = 1;
		}
		vector<int> a, b, c;
		for(int i = 0; i < n; i++) if(!vis[i])
			a.push_back(K + i);
		for(int i = n; i < 2 * n; i++) if(!vis[i])
			b.push_back(K + i);
		for(int i = 2 * n; i < 3 * n; i++) if(!vis[i])
			c.push_back(K + i);
		for(int i = a.size() - 1; i >= 0; i--)
			cout << a[i] << ' ' << b[i] << ' ' << c[i] << "\n";
	}
	else{
		cout << "-1\n";
	}
	return 0;
}
```