#include <bits/stdc++.h>

using namespace std;

struct Node {
    int x, y, z;
    Node() {}
    Node(int _x, int _y, int _z) : x(_x), y(_y), z(_z) {}
};

void solve() {
    int n, k;
    cin >> n >> k;
    if (2*(k-1) > n-1) {
        cout << -1; return;
    }

    int x = k + (n-1)/2, y = k + 2*n-1, z = k + 3*n-1;
    while (x >= k) {
        cout << x << ' ' << y << ' ' << z << "\n";
        x--; y--; z-=2;
    }
    x = k + n-1; z = k + 3*n-2;
    while (y >= k+n) {
        cout << x << ' ' << y << ' ' << z << "\n";
        x--; y--; z-=2;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}

### Prompt

Create a solution in Cpp for 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;

struct Node {
    int x, y, z;
    Node() {}
    Node(int _x, int _y, int _z) : x(_x), y(_y), z(_z) {}
};

void solve() {
    int n, k;
    cin >> n >> k;
    if (2*(k-1) > n-1) {
        cout << -1; return;
    }

    int x = k + (n-1)/2, y = k + 2*n-1, z = k + 3*n-1;
    while (x >= k) {
        cout << x << ' ' << y << ' ' << z << "\n";
        x--; y--; z-=2;
    }
    x = k + n-1; z = k + 3*n-2;
    while (y >= k+n) {
        cout << x << ' ' << y << ' ' << z << "\n";
        x--; y--; z-=2;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}

```