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

int main(){
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  
  ll n,k;
  cin >> n >> k;
  vector<ll>a(n);
  vector<ll>b(n);
  vector<ll>c(n);
  rep(i,n){
    c[i]=k+2*n+i;
  }
  if(2*k-1>n){
    cout<<-1<<endl;
    return 0;
  }
  else{
    if(n%2==1){
      rep(i,n/2+1){
        a[i]=k+2*i;
        b[i]=k+n+n/2-i;
      }
      rep(i,n/2){
        a[i+n/2+1]=k+1+2*i;
        b[i+n/2+1]=k+2*n-1-i;
      }
    }
    else{
      rep(i,n/2){
        a[i]=k+2*i;
        b[i]=k+n+n/2-1-i;
      }
      rep(i,n/2){
        a[i+n/2]=k+1+2*i;
        b[i+n/2]=k+2*n-1-i;
      }
    }
    rep(i,n){
      cout<<a[i]<<" "<<b[i]<<" "<<c[i]<<endl;
    }
  }
}

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

int main(){
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  
  ll n,k;
  cin >> n >> k;
  vector<ll>a(n);
  vector<ll>b(n);
  vector<ll>c(n);
  rep(i,n){
    c[i]=k+2*n+i;
  }
  if(2*k-1>n){
    cout<<-1<<endl;
    return 0;
  }
  else{
    if(n%2==1){
      rep(i,n/2+1){
        a[i]=k+2*i;
        b[i]=k+n+n/2-i;
      }
      rep(i,n/2){
        a[i+n/2+1]=k+1+2*i;
        b[i+n/2+1]=k+2*n-1-i;
      }
    }
    else{
      rep(i,n/2){
        a[i]=k+2*i;
        b[i]=k+n+n/2-1-i;
      }
      rep(i,n/2){
        a[i+n/2]=k+1+2*i;
        b[i+n/2]=k+2*n-1-i;
      }
    }
    rep(i,n){
      cout<<a[i]<<" "<<b[i]<<" "<<c[i]<<endl;
    }
  }
}

```