#include <iostream>
using namespace std;
int main(void){
    long long N,K,k,i;
    cin>>N>>K;
    if(K*2>N+1){
        cout<<-1<<endl;
        return 0;
    }
    long long a[N],b[N],c[N];
    if(N%2==1){
        k=(N+1)/2;
        for(i=0;i<N;i++){
            c[i]=k*5-2+i;
            if(i<k){
                a[i]=k*2-1-i;
            }else{
                a[i]=k*4-2-i;
            }
            b[i]=c[i]-a[i];
        }
        k=k-K;
        for(i=0;i<N;i++){
            a[i]-=k;b[i]-=k;c[i]-=k;
        }
    }else{
        k=N/2;
        for(i=0;i<N;i++){
            c[i]=k*5+i;
            if(i<k){
                a[i]=2*k-1-i;
                b[i]=c[i]-a[i];
            }else{
                a[i]=k*4-1-i;
                b[i]=c[i]-a[i]-1;
            }
        }
        k=k-K;
        for(i=0;i<N;i++){
            a[i]-=k;b[i]-=k;c[i]-=k;
        }
    }
    for(i=0;i<N;i++){
        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 <iostream>
using namespace std;
int main(void){
    long long N,K,k,i;
    cin>>N>>K;
    if(K*2>N+1){
        cout<<-1<<endl;
        return 0;
    }
    long long a[N],b[N],c[N];
    if(N%2==1){
        k=(N+1)/2;
        for(i=0;i<N;i++){
            c[i]=k*5-2+i;
            if(i<k){
                a[i]=k*2-1-i;
            }else{
                a[i]=k*4-2-i;
            }
            b[i]=c[i]-a[i];
        }
        k=k-K;
        for(i=0;i<N;i++){
            a[i]-=k;b[i]-=k;c[i]-=k;
        }
    }else{
        k=N/2;
        for(i=0;i<N;i++){
            c[i]=k*5+i;
            if(i<k){
                a[i]=2*k-1-i;
                b[i]=c[i]-a[i];
            }else{
                a[i]=k*4-1-i;
                b[i]=c[i]-a[i]-1;
            }
        }
        k=k-K;
        for(i=0;i<N;i++){
            a[i]-=k;b[i]-=k;c[i]-=k;
        }
    }
    for(i=0;i<N;i++){
        cout<<a[i]<<" "<<b[i]<<" "<<c[i]<<endl;
    }
}

```