#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)

int main()
{
ll n , k ;
cin >> n >> k ;
if((k+k+2*n-1)*2*n/2>(k+2*n+k+3*n-1)*n/2){
cout << -1 << endl;
return 0;
}
ll a=k,b=(n+1)/2+n+k-1;
ll nu=0;
while(1){
  cout << a << " " << b << " " << k+2*n+nu << endl;
  a+=2;
  b--;
  nu++;
  if(a>=n+k)break;
}
a=k+1;
b=k+2*n-1;
while(1){
  if(a>=n+k)break;
  cout << a << " " << b << " " << k+2*n+nu << endl;
  a+=2;
  b--;
  nu++;
  if(a>=n+k)break;
}
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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)

int main()
{
ll n , k ;
cin >> n >> k ;
if((k+k+2*n-1)*2*n/2>(k+2*n+k+3*n-1)*n/2){
cout << -1 << endl;
return 0;
}
ll a=k,b=(n+1)/2+n+k-1;
ll nu=0;
while(1){
  cout << a << " " << b << " " << k+2*n+nu << endl;
  a+=2;
  b--;
  nu++;
  if(a>=n+k)break;
}
a=k+1;
b=k+2*n-1;
while(1){
  if(a>=n+k)break;
  cout << a << " " << b << " " << k+2*n+nu << endl;
  a+=2;
  b--;
  nu++;
  if(a>=n+k)break;
}
return 0;
}
```