#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <utility>
#include <cmath>
#include <tuple>
#include <queue>
#include <bitset>
#include <set>
#include <map>
#include <list>

using namespace std;


void solve(long long N, long long K){
  if (2*K > N + 1) {
    std::cout << -1 << std::endl;
    return;
  }
  long long M = K + 3*N -1;
  long long a, b, c;
  c = M;
  b = M - N;
  for (int i = 0; i < (N+1)/2; i++) {
    a = K + (N+1)/2 - i -1;
    std::cout << a << " " << b << " " << c << std::endl;
    c -= 2;
    b -= 1;
  }
  c = M - 1;
  for (int i = 0; i < N/2; i++) {
    a = K + (N+1)/2 + N/2 - i-1;
    std::cout << a << " " << b << " " << c << std::endl;
    c -= 2;
    b -= 1;
  }
}

int main(){
  long long N;
  scanf("%lld",&N);
  long long K;
  scanf("%lld",&K);
  solve(N, K);
  return 0;
}

### Prompt

Generate 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 <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <utility>
#include <cmath>
#include <tuple>
#include <queue>
#include <bitset>
#include <set>
#include <map>
#include <list>

using namespace std;


void solve(long long N, long long K){
  if (2*K > N + 1) {
    std::cout << -1 << std::endl;
    return;
  }
  long long M = K + 3*N -1;
  long long a, b, c;
  c = M;
  b = M - N;
  for (int i = 0; i < (N+1)/2; i++) {
    a = K + (N+1)/2 - i -1;
    std::cout << a << " " << b << " " << c << std::endl;
    c -= 2;
    b -= 1;
  }
  c = M - 1;
  for (int i = 0; i < N/2; i++) {
    a = K + (N+1)/2 + N/2 - i-1;
    std::cout << a << " " << b << " " << c << std::endl;
    c -= 2;
    b -= 1;
  }
}

int main(){
  long long N;
  scanf("%lld",&N);
  long long K;
  scanf("%lld",&K);
  solve(N, K);
  return 0;
}

```