#include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T> void drop(const T &x){cout<<x<<endl;exit(0);}

//INSERT ABOVE HERE
signed main(){
  Int n,k;
  cin>>n>>k;
  Int x=3*n/2;
  if(k+x>2*n) drop(-1);

  for(Int i=0,j=0;i<n;i++){
    if(i+x<2*n){
      cout<<k+i<<" "<<k+x+i<<" "<<k+n+n+i*2<<"\n";
    }else{
      cout<<k+i<<" "<<k+n+j<<" "<<k+n+n+j*2+1<<"\n";
      j++;
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T> void drop(const T &x){cout<<x<<endl;exit(0);}

//INSERT ABOVE HERE
signed main(){
  Int n,k;
  cin>>n>>k;
  Int x=3*n/2;
  if(k+x>2*n) drop(-1);

  for(Int i=0,j=0;i<n;i++){
    if(i+x<2*n){
      cout<<k+i<<" "<<k+x+i<<" "<<k+n+n+i*2<<"\n";
    }else{
      cout<<k+i<<" "<<k+n+j<<" "<<k+n+n+j*2+1<<"\n";
      j++;
    }
  }
  return 0;
}

```