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

struct data {
  long long a, b, c;
};

long long n, k;
vector<data> ans;

long long solve() {
  int state = 0, now = 0;
  ans.resize(n);
  for(int i = 0; i < n; ++i) ans[i].c = k + 2 * n + i;
  for(int i = 0; i < 2 * n; ++i) {
    if(state % 2 == 0)
      ans[now].a = k + i;
    else
      ans[now].b = i + k;
    now += 2;
    while(now >= n) {
      ++state;
      now = (state == 1) || (state == 2);
    }
  }
  for(int i = 0; i < n; ++i)
    if(ans[i].a + ans[i].b > ans[i].c) return -1;
  return 0;
}

int main() {
  cin >> n >> k;
  if(solve() == -1)
    cout << -1;
  else
    for(int i = 0; i < n; ++i)
      cout << ans[i].a << " " << ans[i].b << " " << ans[i].c
           << endl;
  return 0;
}
### Prompt

Please create a solution in Cpp 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;

struct data {
  long long a, b, c;
};

long long n, k;
vector<data> ans;

long long solve() {
  int state = 0, now = 0;
  ans.resize(n);
  for(int i = 0; i < n; ++i) ans[i].c = k + 2 * n + i;
  for(int i = 0; i < 2 * n; ++i) {
    if(state % 2 == 0)
      ans[now].a = k + i;
    else
      ans[now].b = i + k;
    now += 2;
    while(now >= n) {
      ++state;
      now = (state == 1) || (state == 2);
    }
  }
  for(int i = 0; i < n; ++i)
    if(ans[i].a + ans[i].b > ans[i].c) return -1;
  return 0;
}

int main() {
  cin >> n >> k;
  if(solve() == -1)
    cout << -1;
  else
    for(int i = 0; i < n; ++i)
      cout << ans[i].a << " " << ans[i].b << " " << ans[i].c
           << endl;
  return 0;
}
```