//07:27~
#include <bits/stdc++.h>
using namespace std;

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    
    int64_t n, k;
    static int64_t a[100000], b[100000], c[100000];
    cin >> n >> k;
    int64_t kmax = (n + 1) / 2;
    if(k > kmax){ cout << -1 << endl; return 0; }
    if(n & 1){
        for(int64_t i=0;i<=n/2;++i){ a[i] = kmax + 2*i; b[i] = 2*kmax+n-1 - i; c[i] = kmax + 2*n + i; }
        for(int64_t i=0;n/2+1+i<n;++i){ a[n/2+1+i] = kmax + 2*i + 1; b[n/2+1+i] = kmax + 2*n - 1 - i; c[n/2+1+i] = kmax + 2*n + n/2+1+i; }
    }else{
        for(int64_t i=0;i<n/2;++i){ a[i] = kmax + 2*i; b[i] = 2*kmax+n-1 - i; c[i] = kmax + 2*n + i; }
        for(int64_t i=0;n/2+i<n;++i){ a[n/2+i] = kmax + 2*i + 1; b[n/2+i] = kmax + 2*n - 1 - i; c[n/2+i] = kmax + 2*n + n/2+i; }
    }
    for(int64_t i=0;i<n;++i){
        cout << a[i] - (kmax - k) << " " << b[i] - (kmax - k) << " " << c[i] - (kmax - k) << endl;
    }
    
    return 0;
}
### Prompt

Your task is to create 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
//07:27~
#include <bits/stdc++.h>
using namespace std;

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    
    int64_t n, k;
    static int64_t a[100000], b[100000], c[100000];
    cin >> n >> k;
    int64_t kmax = (n + 1) / 2;
    if(k > kmax){ cout << -1 << endl; return 0; }
    if(n & 1){
        for(int64_t i=0;i<=n/2;++i){ a[i] = kmax + 2*i; b[i] = 2*kmax+n-1 - i; c[i] = kmax + 2*n + i; }
        for(int64_t i=0;n/2+1+i<n;++i){ a[n/2+1+i] = kmax + 2*i + 1; b[n/2+1+i] = kmax + 2*n - 1 - i; c[n/2+1+i] = kmax + 2*n + n/2+1+i; }
    }else{
        for(int64_t i=0;i<n/2;++i){ a[i] = kmax + 2*i; b[i] = 2*kmax+n-1 - i; c[i] = kmax + 2*n + i; }
        for(int64_t i=0;n/2+i<n;++i){ a[n/2+i] = kmax + 2*i + 1; b[n/2+i] = kmax + 2*n - 1 - i; c[n/2+i] = kmax + 2*n + n/2+i; }
    }
    for(int64_t i=0;i<n;++i){
        cout << a[i] - (kmax - k) << " " << b[i] - (kmax - k) << " " << c[i] - (kmax - k) << endl;
    }
    
    return 0;
}
```