#include <bits/stdc++.h>

using namespace std;

template <class T>inline T updmax(T& a, T b) {return a = max(a, b);}
template <class T>inline T updmin(T& a, T b) {return a = min(a, b);}

class Solution {
public:
	void solve() {
		long long n, k;
		cin >> n >> k;
		if(2*k-1 > n){
			cout << "-1\n";
			return;
		}
		vector<vector<long long>> ans(n, vector<long long>(3, 0));
		for(int i=0; i < n; i++)ans[i][2] = k + 2*n + i;
		int x = k + n;
		for(int i=0; i<n; i+=2, x++)ans[i][1] = x;
		for(int i=1; i<n; i+=2, x++)ans[i][1] = x;
		x = k;
		for(int i=1; i<n; i+=2, x++)ans[i][0] = x;
		for(int i=0; i<n; i+=2, x++)ans[i][0] = x;
		for(int i=0; i < n; i++){
			for(int j=0; j<3; j++)cout << ans[i][j] << " ";
			cout << "\n";
		}
		return;
	}
};

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	Solution solution;
	solution.solve();
	return 0;
}
### Prompt

In cpp, your task is to solve 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;

template <class T>inline T updmax(T& a, T b) {return a = max(a, b);}
template <class T>inline T updmin(T& a, T b) {return a = min(a, b);}

class Solution {
public:
	void solve() {
		long long n, k;
		cin >> n >> k;
		if(2*k-1 > n){
			cout << "-1\n";
			return;
		}
		vector<vector<long long>> ans(n, vector<long long>(3, 0));
		for(int i=0; i < n; i++)ans[i][2] = k + 2*n + i;
		int x = k + n;
		for(int i=0; i<n; i+=2, x++)ans[i][1] = x;
		for(int i=1; i<n; i+=2, x++)ans[i][1] = x;
		x = k;
		for(int i=1; i<n; i+=2, x++)ans[i][0] = x;
		for(int i=0; i<n; i+=2, x++)ans[i][0] = x;
		for(int i=0; i < n; i++){
			for(int j=0; j<3; j++)cout << ans[i][j] << " ";
			cout << "\n";
		}
		return;
	}
};

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	Solution solution;
	solution.solve();
	return 0;
}
```