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

int main() {
	int n,k;cin>>n>>k;

	if(n%2==0){
		if(2*k>n){cout<<-1<<endl;return 0;}
		int m=n/2;
		for(int i=0;i<m;i++){
			cout<<k+2*i<<" "<<k+2*n-1-i<<" "<<k+2*n+m+i<<endl;
		}
		for(int i=0;i<m;i++){
			cout<<k+2*i+1<<" "<<k+n+n-m-1-i<<" "<<k+2*n+i<<endl;
		}
	}
	else{
		if(2*k>n+1){cout<<-1<<endl;return 0;}
		int m=(n+1)/2;

		for(int i=0;i<m;i++){
			cout<<k+2*i<<" "<<k+n+n-1-i<<" "<<k+n+n+m+i-1<<endl;
		}
		for(int i=0;i<m-1;i++){
			cout<<k+2*i+1<<" "<<k+n+n-m-1-i<<" "<<k+2*n+i<<endl;
		}
	}
	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;

int main() {
	int n,k;cin>>n>>k;

	if(n%2==0){
		if(2*k>n){cout<<-1<<endl;return 0;}
		int m=n/2;
		for(int i=0;i<m;i++){
			cout<<k+2*i<<" "<<k+2*n-1-i<<" "<<k+2*n+m+i<<endl;
		}
		for(int i=0;i<m;i++){
			cout<<k+2*i+1<<" "<<k+n+n-m-1-i<<" "<<k+2*n+i<<endl;
		}
	}
	else{
		if(2*k>n+1){cout<<-1<<endl;return 0;}
		int m=(n+1)/2;

		for(int i=0;i<m;i++){
			cout<<k+2*i<<" "<<k+n+n-1-i<<" "<<k+n+n+m+i-1<<endl;
		}
		for(int i=0;i<m-1;i++){
			cout<<k+2*i+1<<" "<<k+n+n-m-1-i<<" "<<k+2*n+i<<endl;
		}
	}
	return 0;
}

```