#include <iostream>
using namespace std;
typedef long long ll;

int main()
{
    ll n, k;
    cin >> n >> k;
    ll a[100002], b[100002], c[100002];
    for(ll i = 0; i < n; i++) c[i] = k + n * 2 + i;
    if(n % 2){
        for(ll i = 0; i <= n / 2; i++) a[i * 2] = k + i, b[i * 2] = k + n + n / 2 + i;
        for(ll i = 0; i < n / 2; i++) a[i * 2 + 1] = k + n / 2 + 1 + i, b[i * 2 + 1] = k + n + i;
    }
    else{
        for(ll i = 0; i < n / 2; i++) a[i * 2] = k + i, b[i * 2] = k + n * 3 / 2 + i;
        for(ll i = 0; i < n / 2 - 1; i++) a[i * 2 + 1] = k + n / 2 + i, b[i * 2 + 1] = k + n + 1 + i;
        a[n - 1] = k + n - 1, b[n - 1] = k + n;
    }
    for(int i = 0; i < n; i++) if(a[i] + b[i] > c[i]){
        cout << -1 << endl;
        return 0;
    }
    for(int i = 0; i < n; i++) cout << a[i] << " " << b[i] << " " << c[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;
typedef long long ll;

int main()
{
    ll n, k;
    cin >> n >> k;
    ll a[100002], b[100002], c[100002];
    for(ll i = 0; i < n; i++) c[i] = k + n * 2 + i;
    if(n % 2){
        for(ll i = 0; i <= n / 2; i++) a[i * 2] = k + i, b[i * 2] = k + n + n / 2 + i;
        for(ll i = 0; i < n / 2; i++) a[i * 2 + 1] = k + n / 2 + 1 + i, b[i * 2 + 1] = k + n + i;
    }
    else{
        for(ll i = 0; i < n / 2; i++) a[i * 2] = k + i, b[i * 2] = k + n * 3 / 2 + i;
        for(ll i = 0; i < n / 2 - 1; i++) a[i * 2 + 1] = k + n / 2 + i, b[i * 2 + 1] = k + n + 1 + i;
        a[n - 1] = k + n - 1, b[n - 1] = k + n;
    }
    for(int i = 0; i < n; i++) if(a[i] + b[i] > c[i]){
        cout << -1 << endl;
        return 0;
    }
    for(int i = 0; i < n; i++) cout << a[i] << " " << b[i] << " " << c[i] << endl;
}
```