#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,int> P;
typedef vector<pair<ll,ll>> vep;
const ll mod=998244353;
const ll INF=1e15;
const int BIG=1e7;
int ans[3][100001];

int main(){
    ll N,K; cin>>N>>K;
    if(2*K>N+1) cout<<-1<<endl;
    else{
        ll M=(N+1)/2;
        for(int i=0;i<M;i++){
            ans[0][i]=M+i;
            ans[1][i]=2*N+i;
            ans[2][i]=M+2*(N+i);
        }
        for(int i=0;i<N-M;i++){
            ans[0][M+i]=2*M+i;
            ans[1][M+i]=N+M+i;
            ans[2][M+i]=M+1+2*(N+i);
        }

        ll X=M-K;
        for(int i=0;i<N;i++){
            cout<<ans[0][i]-X<<' '<<ans[1][i]-X<<' '<<ans[2][i]-X<<endl;
        }
        
    }
}
### 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>
#include <stdlib.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,int> P;
typedef vector<pair<ll,ll>> vep;
const ll mod=998244353;
const ll INF=1e15;
const int BIG=1e7;
int ans[3][100001];

int main(){
    ll N,K; cin>>N>>K;
    if(2*K>N+1) cout<<-1<<endl;
    else{
        ll M=(N+1)/2;
        for(int i=0;i<M;i++){
            ans[0][i]=M+i;
            ans[1][i]=2*N+i;
            ans[2][i]=M+2*(N+i);
        }
        for(int i=0;i<N-M;i++){
            ans[0][M+i]=2*M+i;
            ans[1][M+i]=N+M+i;
            ans[2][M+i]=M+1+2*(N+i);
        }

        ll X=M-K;
        for(int i=0;i<N;i++){
            cout<<ans[0][i]-X<<' '<<ans[1][i]-X<<' '<<ans[2][i]-X<<endl;
        }
        
    }
}
```