#include <bits/stdc++.h>
#define rep(i,l,r) for (int i=l,i##end=r;i<=i##end;++i)
#define per(i,r,l) for (int i=r,i##end=l;i>=i##end;--i)
#define ll long long
#define pr pair<int,int>
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int N=1e6+6;
int main(){
    int n,l; cin>>n>>l;
    if (2*l>n+1){ cout<<-1<<endl; return 0; }
    if (n&1){
        int t=n+1>>1;
        rep(i,1,t) printf("%d %d %d\n",l+2*(i-1),l+3*t-1-i,l+4*t-3+i);
        rep(i,1,t-1) printf("%d %d %d\n",l+1+2*(i-1),l+4*t-2-i,l+5*t-3+i);
    }
    else{
        int t=n>>1;
        rep(i,1,t) printf("%d %d %d\n",l+2*(i-1),l+3*t-i,l+4*t-1+i);
        rep(i,1,t) printf("%d %d %d\n",l-1+2*i,l+4*t-i,l+5*t-1+i);
    }
    return 0;
}

### Prompt

Please create a solution in cpp 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>
#define rep(i,l,r) for (int i=l,i##end=r;i<=i##end;++i)
#define per(i,r,l) for (int i=r,i##end=l;i>=i##end;--i)
#define ll long long
#define pr pair<int,int>
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int N=1e6+6;
int main(){
    int n,l; cin>>n>>l;
    if (2*l>n+1){ cout<<-1<<endl; return 0; }
    if (n&1){
        int t=n+1>>1;
        rep(i,1,t) printf("%d %d %d\n",l+2*(i-1),l+3*t-1-i,l+4*t-3+i);
        rep(i,1,t-1) printf("%d %d %d\n",l+1+2*(i-1),l+4*t-2-i,l+5*t-3+i);
    }
    else{
        int t=n>>1;
        rep(i,1,t) printf("%d %d %d\n",l+2*(i-1),l+3*t-i,l+4*t-1+i);
        rep(i,1,t) printf("%d %d %d\n",l-1+2*i,l+4*t-i,l+5*t-1+i);
    }
    return 0;
}

```