#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;
#define REP(i,n) for (ll i = 0; i < (n); ++i)


int main(){
    ll n, k;
    cin >> n >> k;
    if(2 * k > n + 1){
        cout << -1 << endl;
        return 0;
    }

    const ll s = (n + 1) / 2;

    vector<ll> a(n), b(n), c(n);
    REP(i, n){
        c.at(i) = k + 2 * n + i;
    }

    for(ll i = 0; i < s; ++i){
        a.at(i) = k + 2 * i;
        b.at(i) = k + s + n - i - 1;
    }
    for(ll i = 0; i < n - s; ++i){
        a.at(s + i) = k + 2 * i + 1;
        b.at(s + i) = k + 2 * n - 1 - i;
    }

    REP(i, n){
        cout << a.at(i) << ' ' << b.at(i) << ' ' << c.at(i) << endl;
    }
    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 <iostream>
#include <vector>

using namespace std;
typedef long long ll;
#define REP(i,n) for (ll i = 0; i < (n); ++i)


int main(){
    ll n, k;
    cin >> n >> k;
    if(2 * k > n + 1){
        cout << -1 << endl;
        return 0;
    }

    const ll s = (n + 1) / 2;

    vector<ll> a(n), b(n), c(n);
    REP(i, n){
        c.at(i) = k + 2 * n + i;
    }

    for(ll i = 0; i < s; ++i){
        a.at(i) = k + 2 * i;
        b.at(i) = k + s + n - i - 1;
    }
    for(ll i = 0; i < n - s; ++i){
        a.at(s + i) = k + 2 * i + 1;
        b.at(s + i) = k + 2 * n - 1 - i;
    }

    REP(i, n){
        cout << a.at(i) << ' ' << b.at(i) << ' ' << c.at(i) << endl;
    }
    return 0;
}

```