#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;
const int INF=1e9;
const ll LINF=1e18;

int main(){
    int n,k;
    cin>>n>>k;
    if(2*k-1>n){
        cout<<-1<<endl;
        return 0;
    }
    vector<pair<int,pair<int,int>>> p;
    for(int i=0;i<(n+1)/2;i++){
        int a=k+i;
        int b=k+n+n-(n+1)/2+i;
        p.push_back({a+b,{a,b}});
    }
    for(int i=0;i<n-(n+1)/2;i++){
        int a=k+(n+1)/2+i;
        int b=k+n+i;
        p.push_back({a+b,{a,b}});
    }
    sort(p.begin(),p.end());
    for(int i=0;i<n;i++){
        cout<<p[i].second.first<<" "<<p[i].second.second<<" "<<k+2*n+i<<endl;
    }
}
### Prompt

Your challenge is to write a cpp solution 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>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;
const int INF=1e9;
const ll LINF=1e18;

int main(){
    int n,k;
    cin>>n>>k;
    if(2*k-1>n){
        cout<<-1<<endl;
        return 0;
    }
    vector<pair<int,pair<int,int>>> p;
    for(int i=0;i<(n+1)/2;i++){
        int a=k+i;
        int b=k+n+n-(n+1)/2+i;
        p.push_back({a+b,{a,b}});
    }
    for(int i=0;i<n-(n+1)/2;i++){
        int a=k+(n+1)/2+i;
        int b=k+n+i;
        p.push_back({a+b,{a,b}});
    }
    sort(p.begin(),p.end());
    for(int i=0;i<n;i++){
        cout<<p[i].second.first<<" "<<p[i].second.second<<" "<<k+2*n+i<<endl;
    }
}
```