#include <bits/stdc++.h>
using namespace std;
using Int = int_fast64_t;

Int n, k;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n >> k;
	Int sl = 0, sr = 0;
	for(Int i=0; i<2*n; ++i)
		sl += k+i;
	for(Int i=2*n; i<3*n; ++i)
		sr += k+i;
	if(sl > sr){
		cout << -1 << "\n";
		return 0;
	}
	vector<Int> a(n), b(n), c(n);
	for(Int i=0; i<n; ++i)
		c[i] = k+3*n-1-i;
	Int cnt = 0;
	for(Int i=0; i<n; i+=2){
		b[i] = k+2*n-1-cnt;
		++cnt;
	}
	for(Int i=1; i<n; i+=2){
		b[i] = k+2*n-1-cnt;
		++cnt;
	}
	cnt = 0;
	for(Int i=1; i<n; i+=2){
		a[i] = k+n-1-cnt;
		++cnt;
	}
	for(Int i=0; i<n; i+=2){
		a[i] = k+n-1-cnt;
		++cnt;
	}
	for(Int i=0; i<n; ++i)
		cout << a[i] << " " << b[i] << " " << c[i] << "\n";
}
### Prompt

Create a solution in cpp for 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;
using Int = int_fast64_t;

Int n, k;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n >> k;
	Int sl = 0, sr = 0;
	for(Int i=0; i<2*n; ++i)
		sl += k+i;
	for(Int i=2*n; i<3*n; ++i)
		sr += k+i;
	if(sl > sr){
		cout << -1 << "\n";
		return 0;
	}
	vector<Int> a(n), b(n), c(n);
	for(Int i=0; i<n; ++i)
		c[i] = k+3*n-1-i;
	Int cnt = 0;
	for(Int i=0; i<n; i+=2){
		b[i] = k+2*n-1-cnt;
		++cnt;
	}
	for(Int i=1; i<n; i+=2){
		b[i] = k+2*n-1-cnt;
		++cnt;
	}
	cnt = 0;
	for(Int i=1; i<n; i+=2){
		a[i] = k+n-1-cnt;
		++cnt;
	}
	for(Int i=0; i<n; i+=2){
		a[i] = k+n-1-cnt;
		++cnt;
	}
	for(Int i=0; i<n; ++i)
		cout << a[i] << " " << b[i] << " " << c[i] << "\n";
}
```