#include <bits/stdc++.h>
using namespace std;

typedef pair <int, int> ii;
typedef pair <int, ii> iii;

int n, k;
vector <iii> seq;

int main()
{
    scanf("%d %d", &n, &k);
    if (k > (n + 1) / 2) printf("-1\n");
    else {
        int fir = n - (n + 1) / 2;
        for (int i = 0; i < n; i++) {
            seq.push_back(iii(i + n + fir, ii(i, n + fir)));
            fir = (fir + 1) % n;
        }
        sort(seq.begin(), seq.end());
        for (int i = 0; i < seq.size(); i++)
            printf("%d %d %d\n", k + seq[i].second.first, k + seq[i].second.second, k + 2 * n + i);
    }
    return 0;
}

### Prompt

Please create a solution in cpp 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
#include <bits/stdc++.h>
using namespace std;

typedef pair <int, int> ii;
typedef pair <int, ii> iii;

int n, k;
vector <iii> seq;

int main()
{
    scanf("%d %d", &n, &k);
    if (k > (n + 1) / 2) printf("-1\n");
    else {
        int fir = n - (n + 1) / 2;
        for (int i = 0; i < n; i++) {
            seq.push_back(iii(i + n + fir, ii(i, n + fir)));
            fir = (fir + 1) % n;
        }
        sort(seq.begin(), seq.end());
        for (int i = 0; i < seq.size(); i++)
            printf("%d %d %d\n", k + seq[i].second.first, k + seq[i].second.second, k + 2 * n + i);
    }
    return 0;
}

```