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

int main() {
    int n, k; cin >> n >> k;
    vector<pair<int, pair<int, int>>> d;
    vector<int> a(n), b(n), c(n);
    for (int i = 0; i < n/2; i++) {
        d.push_back({i+1, {i, i*2+1}});
        d.push_back({i-n/2, {i+n/2, i*2}});
    }
    if (n & 1) {
        d.push_back({0, {n-1, n-1}});
    }
    sort(d.begin(), d.end());
    for (int i = 0; i < n; i++) {
        auto u = n+k+i;
        auto p = d[i].second.first + k;
        auto q = d[i].second.second + k + n*2;
        if (u > q-p) {
            cout << -1 << endl;
            return 0;
        }
        a[i] = p;
        b[i] = u;
        c[i] = q;
    }
    for (int i = 0; i < n; i++) {
        cout << a[i] << " " << b[i] << " " << c[i] << endl;
    }
    return 0;
}
### 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() {
    int n, k; cin >> n >> k;
    vector<pair<int, pair<int, int>>> d;
    vector<int> a(n), b(n), c(n);
    for (int i = 0; i < n/2; i++) {
        d.push_back({i+1, {i, i*2+1}});
        d.push_back({i-n/2, {i+n/2, i*2}});
    }
    if (n & 1) {
        d.push_back({0, {n-1, n-1}});
    }
    sort(d.begin(), d.end());
    for (int i = 0; i < n; i++) {
        auto u = n+k+i;
        auto p = d[i].second.first + k;
        auto q = d[i].second.second + k + n*2;
        if (u > q-p) {
            cout << -1 << endl;
            return 0;
        }
        a[i] = p;
        b[i] = u;
        c[i] = q;
    }
    for (int i = 0; i < n; i++) {
        cout << a[i] << " " << b[i] << " " << c[i] << endl;
    }
    return 0;
}
```