#include<iostream>
using namespace std;
#define ll long long
#define int long long

pair<int,int> ans[100010];
int ans2[100010];
int p;

signed main(){
    int n,k;cin>>n>>k;

    if(k>(n+1)/2){
        cout<<"-1"<<endl;
        return 0;
    }

    ll now=(3*n+1)/2;
    if(n%2){
        for(int i=1;i<=n;i++){
            int tmp=i%2==0?(n+1)/2+i/2:(i+1)/2;
            ans[p++]={tmp,now+i-tmp};
        }
    }
    else{
        for(int i=1;i<n;i++){
            int tmp=i%2==0?n/2+1+i/2:(i+1)/2;
            ans[p++]={tmp,(now+1+i/2*2)-tmp};
        }
        ans[p++]={n/2+1,2*n};
    }

    for(int i=0;i<n;i++)ans2[i]=k+n*2+i,ans[i].first+=k-1,ans[i].second+=k-1;
    for(int i=0;i<n;i++){
        if(ans[i].first+ans[i].second>ans2[i]){
            cout<<"-1"<<endl;
            return 0;
        }
    }
    for(int i=0;i<n;i++){
        cout<<ans[i].first<<" "<<ans[i].second<<" "<<ans2[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;
#define ll long long
#define int long long

pair<int,int> ans[100010];
int ans2[100010];
int p;

signed main(){
    int n,k;cin>>n>>k;

    if(k>(n+1)/2){
        cout<<"-1"<<endl;
        return 0;
    }

    ll now=(3*n+1)/2;
    if(n%2){
        for(int i=1;i<=n;i++){
            int tmp=i%2==0?(n+1)/2+i/2:(i+1)/2;
            ans[p++]={tmp,now+i-tmp};
        }
    }
    else{
        for(int i=1;i<n;i++){
            int tmp=i%2==0?n/2+1+i/2:(i+1)/2;
            ans[p++]={tmp,(now+1+i/2*2)-tmp};
        }
        ans[p++]={n/2+1,2*n};
    }

    for(int i=0;i<n;i++)ans2[i]=k+n*2+i,ans[i].first+=k-1,ans[i].second+=k-1;
    for(int i=0;i<n;i++){
        if(ans[i].first+ans[i].second>ans2[i]){
            cout<<"-1"<<endl;
            return 0;
        }
    }
    for(int i=0;i<n;i++){
        cout<<ans[i].first<<" "<<ans[i].second<<" "<<ans2[i]<<endl;
    }
}

```