#include <bits/stdc++.h>
using namespace std;
using Int = long long;
int main() {
    int N, K; cin >> N >> K;
    vector<int> A(N), B(N), C(N);
    int t = K;
    for (int i = 0; i < N; i += 2) A[i] = t++;
    for (int i = 1; i < N; i += 2) A[i] = t++;
    for (int i = 1; i < N; i += 2) B[i] = t++;
    for (int i = 0; i < N; i += 2) B[i] = t++;
    for (int i = 0; i < N; i += 1) C[i] = t++;
    for (int i = 0; i < N; i++) {
        if (A[i] + B[i] > C[i]) {
            cout << "-1\n";
            return 0;
        }
    }
    for (int i = 0; i < N; i++) {
        cout << A[i] << ' ' << B[i] << ' ' << C[i] << '\n';
    }
    return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
using Int = long long;
int main() {
    int N, K; cin >> N >> K;
    vector<int> A(N), B(N), C(N);
    int t = K;
    for (int i = 0; i < N; i += 2) A[i] = t++;
    for (int i = 1; i < N; i += 2) A[i] = t++;
    for (int i = 1; i < N; i += 2) B[i] = t++;
    for (int i = 0; i < N; i += 2) B[i] = t++;
    for (int i = 0; i < N; i += 1) C[i] = t++;
    for (int i = 0; i < N; i++) {
        if (A[i] + B[i] > C[i]) {
            cout << "-1\n";
            return 0;
        }
    }
    for (int i = 0; i < N; i++) {
        cout << A[i] << ' ' << B[i] << ' ' << C[i] << '\n';
    }
    return 0;
}

```