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

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, k;
    cin >> n >> k;
    if (2 * k > n + 1) {
        return cout << -1, 0;
    } else {
        vector<pair<int, int>> ve;
        for (int i = 0; i < n; i++) {
            if (i < (n + 1) / 2) {
                ve.push_back({k + i, k + 2 * n - 1 - 2 * i});
            } else {
                ve.push_back({k + i, k + 2 * n - 2 - 2 * (i - (n + 1) / 2)});
            }
        }
        sort(ve.begin(), ve.end(), [](const pair<int, int> &u, const pair<int, int> &v) {
            return u.first + u.second < v.first + v.second;
        });
        for (int i = 0; i < n; i++) {
            cout << ve[i].first << " " << ve[i].second << " " << k + 2 * n + i << '\n';
        }
    }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, k;
    cin >> n >> k;
    if (2 * k > n + 1) {
        return cout << -1, 0;
    } else {
        vector<pair<int, int>> ve;
        for (int i = 0; i < n; i++) {
            if (i < (n + 1) / 2) {
                ve.push_back({k + i, k + 2 * n - 1 - 2 * i});
            } else {
                ve.push_back({k + i, k + 2 * n - 2 - 2 * (i - (n + 1) / 2)});
            }
        }
        sort(ve.begin(), ve.end(), [](const pair<int, int> &u, const pair<int, int> &v) {
            return u.first + u.second < v.first + v.second;
        });
        for (int i = 0; i < n; i++) {
            cout << ve[i].first << " " << ve[i].second << " " << k + 2 * n + i << '\n';
        }
    }
}

```