#include <bits/stdc++.h>
#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;
using ll = long long;
ll inf = 100000000007ll;

int main(void)
{
  ll n,k;
  cin >> n >> k;
  vector<ll> a(n),b(n);
  vector<tuple<ll,ll,ll> > sab(n);
  REP(i,n) b[i] = k+2*n-1 - i;
  REP(i,n) a[i] = k+((i+n)%2 ? (i+n)/2 : i/2);
  REP(i,n) sab[i] = make_tuple(a[i]+b[i],a[i],b[i]);
  sort(sab.begin(),sab.end());
  ll ma;
  tie(ma,ignore,ignore) = sab.back();
  if(ma >= k+3*n) {
    cout << -1 << endl;
    return 0;
  }
  REP(i,n) {
    ll x,y;
    tie(ignore,x,y) = sab[i];
    if(x+y>k+2*n+i) {
      cout << -1 << endl;
      return 0;
    }
    cout << x << " " << y << " " << k+2*n+i << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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 <bits/stdc++.h>
#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;
using ll = long long;
ll inf = 100000000007ll;

int main(void)
{
  ll n,k;
  cin >> n >> k;
  vector<ll> a(n),b(n);
  vector<tuple<ll,ll,ll> > sab(n);
  REP(i,n) b[i] = k+2*n-1 - i;
  REP(i,n) a[i] = k+((i+n)%2 ? (i+n)/2 : i/2);
  REP(i,n) sab[i] = make_tuple(a[i]+b[i],a[i],b[i]);
  sort(sab.begin(),sab.end());
  ll ma;
  tie(ma,ignore,ignore) = sab.back();
  if(ma >= k+3*n) {
    cout << -1 << endl;
    return 0;
  }
  REP(i,n) {
    ll x,y;
    tie(ignore,x,y) = sab[i];
    if(x+y>k+2*n+i) {
      cout << -1 << endl;
      return 0;
    }
    cout << x << " " << y << " " << k+2*n+i << endl;
  }
  return 0;
}

```