#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef pair<int,int>P;
typedef long long ll;

int a[200000],b[200000],c[200000];

int main(){
	int n,K;cin>>n>>K;
	if(K*2>n+1){
		puts("-1");return 0;
	}
	rep(i,n)c[i]=K+2*n+i;
	set<int>se;
	rep(i,2*n)se.insert(K+i);
	int A=K,B=K+2*n-1;
	for(int i=K-1;i>=0;i--){
		a[i]=A;se.erase(A);
		b[i]=B;se.erase(B);
		A++;
		B-=2;
	}
	for(int i=n-1;i>=K;i--){
		a[i]=A;se.erase(A);
		b[i]=*se.rbegin();se.erase(--se.end());
		A++;
	}
	rep(i,n){
		printf("%d %d %d\n",a[i],b[i],c[i]);
	}
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef pair<int,int>P;
typedef long long ll;

int a[200000],b[200000],c[200000];

int main(){
	int n,K;cin>>n>>K;
	if(K*2>n+1){
		puts("-1");return 0;
	}
	rep(i,n)c[i]=K+2*n+i;
	set<int>se;
	rep(i,2*n)se.insert(K+i);
	int A=K,B=K+2*n-1;
	for(int i=K-1;i>=0;i--){
		a[i]=A;se.erase(A);
		b[i]=B;se.erase(B);
		A++;
		B-=2;
	}
	for(int i=n-1;i>=K;i--){
		a[i]=A;se.erase(A);
		b[i]=*se.rbegin();se.erase(--se.end());
		A++;
	}
	rep(i,n){
		printf("%d %d %d\n",a[i],b[i],c[i]);
	}
}
```