#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
typedef pair<int,P> IP;
vector<IP> ans,fans;
template<typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}

signed main(){
  int n,k;cin>>n>>k;
  vector<int> v(3*n);
  for(int i=0;i<3*n;i++)v[i]=i+k;
  int r=2*n-1;
  for(int i=0;i<n;i+=2)ans.push_back(IP(v[i]+v[r],P(v[i],v[r]))),r--;
  for(int i=1;i<n;i+=2)ans.push_back(IP(v[i]+v[r],P(v[i],v[r]))),r--;
  sort(ans.rbegin(),ans.rend());
  for(int i=0;i<n;i++){
    fans.push_back(IP(ans[i].second.first,P(ans[i].second.second,v[3*n-1-i])));
    if(fans[i].first+fans[i].second.first>fans[i].second.second)fin(-1);
  }
  for(auto p:fans)cout<<p.first<<" "<<p.second.first<<" "<<p.second.second<<endl;
}

### 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;
typedef pair<int,int> P;
typedef pair<int,P> IP;
vector<IP> ans,fans;
template<typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}

signed main(){
  int n,k;cin>>n>>k;
  vector<int> v(3*n);
  for(int i=0;i<3*n;i++)v[i]=i+k;
  int r=2*n-1;
  for(int i=0;i<n;i+=2)ans.push_back(IP(v[i]+v[r],P(v[i],v[r]))),r--;
  for(int i=1;i<n;i+=2)ans.push_back(IP(v[i]+v[r],P(v[i],v[r]))),r--;
  sort(ans.rbegin(),ans.rend());
  for(int i=0;i<n;i++){
    fans.push_back(IP(ans[i].second.first,P(ans[i].second.second,v[3*n-1-i])));
    if(fans[i].first+fans[i].second.first>fans[i].second.second)fin(-1);
  }
  for(auto p:fans)cout<<p.first<<" "<<p.second.first<<" "<<p.second.second<<endl;
}

```