#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
typedef long long ll;
typedef pair<ll, ll> P;
int main()
{
    ll n;
    ll k;
    cin >> n >> k;
    vector<ll> a(n), b(n), c(n);
    vector<P> res(n);
    rep(i, n)
    {
        c[i] = 2 * n + k + i;
        a[i] = k + i;
        b[i] = n + k + i;
    }
    reverse(b.begin(), b.end());
    int harf = n/2;
    for(int i = 0;i<harf-i && harf-i<n;i++){
        swap(a[i],a[harf-i]);
    }
    for(int i = 0;harf+1+i<n-1-i;i++){
        swap(a[i+harf+1],a[n-1-i]);
    }
    rep(i,n){
        res[i] = P(a[i]+b[i],a[i]);
    }
    sort(res.begin(), res.end());
    rep(i,n){
        if(res[i].first>c[i]){
            cout << -1 << endl;
            return 0;
        }
    }
    rep(i, n)
    {
        cout << res[i].second << ' ' << res[i].first - res[i].second << ' ' << c[i] << endl;
    }
    return 0;
}
### 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;
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
typedef long long ll;
typedef pair<ll, ll> P;
int main()
{
    ll n;
    ll k;
    cin >> n >> k;
    vector<ll> a(n), b(n), c(n);
    vector<P> res(n);
    rep(i, n)
    {
        c[i] = 2 * n + k + i;
        a[i] = k + i;
        b[i] = n + k + i;
    }
    reverse(b.begin(), b.end());
    int harf = n/2;
    for(int i = 0;i<harf-i && harf-i<n;i++){
        swap(a[i],a[harf-i]);
    }
    for(int i = 0;harf+1+i<n-1-i;i++){
        swap(a[i+harf+1],a[n-1-i]);
    }
    rep(i,n){
        res[i] = P(a[i]+b[i],a[i]);
    }
    sort(res.begin(), res.end());
    rep(i,n){
        if(res[i].first>c[i]){
            cout << -1 << endl;
            return 0;
        }
    }
    rep(i, n)
    {
        cout << res[i].second << ' ' << res[i].first - res[i].second << ' ' << c[i] << endl;
    }
    return 0;
}
```