#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
using T = tuple<ll,ll,ll>;
#define chmax(x,y) x = max(x,y)
#define chmin(x,y) x = min(x,y)

ll n,k;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);

cin>>n>>k;
vector<T>ans(n);
vector<ll>a(n);
vector<ll>b(n);
    ll cur=k;
    for(int i=0;i<n;i+=2){
        a[i]=cur++;
    }
    for(int i=1;i<n;i+=2){
        a[i]=cur++;
    }
    for(int i=1;i<n;i+=2){
        b[i]=cur++;
    }
    for(int i=0;i<n;i+=2){
        b[i]=cur++;
    } 
    rep(i,n){
        ll c=k+2*n+i;
       if(a[i]+b[i]<=c)ans[i]=(T(a[i],b[i],c));
       else{
           cout<<-1<<endl;
           return 0;
       }
    } 
    rep(i,n){
        T tp=ans[i];
        ll f,s,t;
        tie(f,s,t)=tp;
        cout<<f<<" "<<s<<" "<<t<<endl;
    }
return 0;
}
### 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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
using T = tuple<ll,ll,ll>;
#define chmax(x,y) x = max(x,y)
#define chmin(x,y) x = min(x,y)

ll n,k;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);

cin>>n>>k;
vector<T>ans(n);
vector<ll>a(n);
vector<ll>b(n);
    ll cur=k;
    for(int i=0;i<n;i+=2){
        a[i]=cur++;
    }
    for(int i=1;i<n;i+=2){
        a[i]=cur++;
    }
    for(int i=1;i<n;i+=2){
        b[i]=cur++;
    }
    for(int i=0;i<n;i+=2){
        b[i]=cur++;
    } 
    rep(i,n){
        ll c=k+2*n+i;
       if(a[i]+b[i]<=c)ans[i]=(T(a[i],b[i],c));
       else{
           cout<<-1<<endl;
           return 0;
       }
    } 
    rep(i,n){
        T tp=ans[i];
        ll f,s,t;
        tie(f,s,t)=tp;
        cout<<f<<" "<<s<<" "<<t<<endl;
    }
return 0;
}
```