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

struct ab{int a; int b;};
bool cmp(ab A, ab B){
    int Asum = A.a + A.b;
    int Bsum = B.a + B.b;
    return Asum < Bsum;
}

// (K-1)を引いた状態でとりあえず解を作る

int main() {
    int N, K;
    cin >> N >> K;
    vector<ab>ans;

    // [1, 2N]から生成
    int L = 1;
    int R;
    while (L <= N){
        R = 2*N - L/2;
        ans.push_back({L, R});
        L += 2;
    }
    L = 2;
    R--;
    while (L <= N){
        ans.push_back({L, R});
        R--;
        L += 2;
    }

    sort(ans.begin(), ans.end(), cmp);

    int min = ans[0].a + ans[0].b;
    min += (K - 1) * 2;
    if((K + 2*N) < min){
        cout << -1 << "\n";
        return 0;
    }

    for (int i = 0; i < N; ++i) {
        int a = ans[i].a + K - 1;
        int b = ans[i].b + K - 1;
        cout << a << " ";
        cout << b << " ";
        cout << K + 2*N + i << "\n";
    }
    return 0;
}
### Prompt

Your task is to create a Cpp solution 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;
using ll = long long;

struct ab{int a; int b;};
bool cmp(ab A, ab B){
    int Asum = A.a + A.b;
    int Bsum = B.a + B.b;
    return Asum < Bsum;
}

// (K-1)を引いた状態でとりあえず解を作る

int main() {
    int N, K;
    cin >> N >> K;
    vector<ab>ans;

    // [1, 2N]から生成
    int L = 1;
    int R;
    while (L <= N){
        R = 2*N - L/2;
        ans.push_back({L, R});
        L += 2;
    }
    L = 2;
    R--;
    while (L <= N){
        ans.push_back({L, R});
        R--;
        L += 2;
    }

    sort(ans.begin(), ans.end(), cmp);

    int min = ans[0].a + ans[0].b;
    min += (K - 1) * 2;
    if((K + 2*N) < min){
        cout << -1 << "\n";
        return 0;
    }

    for (int i = 0; i < N; ++i) {
        int a = ans[i].a + K - 1;
        int b = ans[i].b + K - 1;
        cout << a << " ";
        cout << b << " ";
        cout << K + 2*N + i << "\n";
    }
    return 0;
}
```