#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MOD = 1000000007;
 //pair<pair<int,int>,int> p[100001];
signed main() {
 int n,k;
 cin>>n>>k;
 if(2*k>n+1){
   cout<<-1;
   return 0;
 }
 if(n%2==1){
   int t=(n+1)/2 -k;
   int a=n,b=n+(n+1)/2;
   int bo=b;
   for(int i=0;i<n;i++){
     cout<<a-t<<' '<<b-t<<' '<<a+b-t<<endl;
     a++;b++;
     if(a==bo)a-=n;
   }
 }else{
   int t=n/2 -k;
   int a=n,b=n+n/2;
   int bo=b;
   for(int i=0;i<n/2;i++){
     cout<<a-t<<' '<<b-t<<' '<<a+b-t<<endl;
     a++;b++;
     if(a==bo)a-=n;
 }
 for(int i=0;i<n/2;i++){
     cout<<a-t<<' '<<b-t<<' '<<a+b-t+1<<endl;
     a++;b++;
     if(a==bo)a-=n;
 }
 }
return 0;
}

### Prompt

Create a solution in Cpp for 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;
#define int long long
const int MOD = 1000000007;
 //pair<pair<int,int>,int> p[100001];
signed main() {
 int n,k;
 cin>>n>>k;
 if(2*k>n+1){
   cout<<-1;
   return 0;
 }
 if(n%2==1){
   int t=(n+1)/2 -k;
   int a=n,b=n+(n+1)/2;
   int bo=b;
   for(int i=0;i<n;i++){
     cout<<a-t<<' '<<b-t<<' '<<a+b-t<<endl;
     a++;b++;
     if(a==bo)a-=n;
   }
 }else{
   int t=n/2 -k;
   int a=n,b=n+n/2;
   int bo=b;
   for(int i=0;i<n/2;i++){
     cout<<a-t<<' '<<b-t<<' '<<a+b-t<<endl;
     a++;b++;
     if(a==bo)a-=n;
 }
 for(int i=0;i<n/2;i++){
     cout<<a-t<<' '<<b-t<<' '<<a+b-t+1<<endl;
     a++;b++;
     if(a==bo)a-=n;
 }
 }
return 0;
}

```