#include <bits/stdc++.h>
using namespace std;

int main() {
  int N,K;
  cin >> N >> K;
  vector<int> A(N),B(N),C(N);
  
  for(int i=0;i<N;i+=2){
    A[i] = K;
    K++;
  }
  for(int i=1;i<N;i+=2){
    A[i] = K;
    K++;
  }
  for(int i=1;i<N;i+=2){
    B[i] = K;
    K++;
  }
  for(int i=0;i<N;i+=2){
    B[i] = K;
    K++;
  }
  for(int i=0;i<N;i++){
    C[i] = K;
    K++;
  }
  
  for(int i=0;i<N;i++){
    if(A[i]+B[i]>C[i]){
      cout << -1 << endl;
      return 0;
    }
  }
  for(int i=0;i<N;i++) cout << A[i] << " " << B[i] << " " << C[i] << endl;
  return 0;
}

### 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;

int main() {
  int N,K;
  cin >> N >> K;
  vector<int> A(N),B(N),C(N);
  
  for(int i=0;i<N;i+=2){
    A[i] = K;
    K++;
  }
  for(int i=1;i<N;i+=2){
    A[i] = K;
    K++;
  }
  for(int i=1;i<N;i+=2){
    B[i] = K;
    K++;
  }
  for(int i=0;i<N;i+=2){
    B[i] = K;
    K++;
  }
  for(int i=0;i<N;i++){
    C[i] = K;
    K++;
  }
  
  for(int i=0;i<N;i++){
    if(A[i]+B[i]>C[i]){
      cout << -1 << endl;
      return 0;
    }
  }
  for(int i=0;i<N;i++) cout << A[i] << " " << B[i] << " " << C[i] << endl;
  return 0;
}

```