#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;++i)
#define debug(x) cerr<<#x<<": "<<x<<endl
#define all(x) x.begin(), x.end()
using ll = long long;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n, k;
    cin >> n >> k;
    if(2*k-1>n) cout<<"-1\n";
    else {
        ll a = n/2, b=2*n-a;
        vector<tuple<ll,ll,ll>> x;
        auto app = [&](ll u, ll v) { x.emplace_back(u+v,u,v); };
        rep(i,0,a) app(k+i,k+b+i);
        rep(i,0,n-a) app(k+a+i,k+n+i);
        sort(all(x));
        rep(i,0,n) {
            ll w,u,v=k+2*n+i;
            tie(ignore,w,u) = x[i];
            cout<<w<<" "<<u<<" "<<v<<endl;
        }
    }
}

### 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;
#define rep(i,a,b) for(int i=a;i<b;++i)
#define debug(x) cerr<<#x<<": "<<x<<endl
#define all(x) x.begin(), x.end()
using ll = long long;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n, k;
    cin >> n >> k;
    if(2*k-1>n) cout<<"-1\n";
    else {
        ll a = n/2, b=2*n-a;
        vector<tuple<ll,ll,ll>> x;
        auto app = [&](ll u, ll v) { x.emplace_back(u+v,u,v); };
        rep(i,0,a) app(k+i,k+b+i);
        rep(i,0,n-a) app(k+a+i,k+n+i);
        sort(all(x));
        rep(i,0,n) {
            ll w,u,v=k+2*n+i;
            tie(ignore,w,u) = x[i];
            cout<<w<<" "<<u<<" "<<v<<endl;
        }
    }
}

```