#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

const int mod = 998244353;

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

    vector<int> x(n), y(n), z(n);

    for(int i = 0; i < n; i++){
        if(i&1){
            x[i] = k + ((n+1)/2) + i/2;
            y[i] = k + n + i/2;
        }else{
            x[i] = k + i/2;
            y[i] = k + n + n/2 + i/2;
        }
        z[i] = k + 2*n + i;
    }
    for(int i = 0; i < n; i++){
        printf("%d %d %d\n", x[i], y[i], z[i]);
    }
}

### 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;
using ll = long long;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

const int mod = 998244353;

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

    vector<int> x(n), y(n), z(n);

    for(int i = 0; i < n; i++){
        if(i&1){
            x[i] = k + ((n+1)/2) + i/2;
            y[i] = k + n + i/2;
        }else{
            x[i] = k + i/2;
            y[i] = k + n + n/2 + i/2;
        }
        z[i] = k + 2*n + i;
    }
    for(int i = 0; i < n; i++){
        printf("%d %d %d\n", x[i], y[i], z[i]);
    }
}

```