#include <bits/stdc++.h>

using namespace std;

int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  set<int> st;
  for (int i = 0; i < n; ++i) {
    st.insert(k + i);
  }
  vector<tuple<int, int, int>> ans;
  int c = k + 3 * n - 1;
  for (int i = n - 1; i >= 0; i -= 2) {
    int b = k + n + i;
    auto it = st.lower_bound(c - b + 1);
    if (it == st.begin()) {
      puts("-1");
      return 0;
    }
    --it;
    int a = *it;
    st.erase(it);
    ans.emplace_back(a, b, c);
    --c;
  }
  for (int i = n - 2; i >= 0; i -= 2) {
    int b = k + n + i;
    auto it = st.lower_bound(c - b + 1);
    if (it == st.begin()) {
      puts("-1");
      return 0;
    }
    --it;
    int a = *it;
    st.erase(it);
    ans.emplace_back(a, b, c);
    --c;
  }
  for (auto it : ans) {
    int a, b, c;
    tie(a, b, c) = it;
    printf("%d %d %d\n", a, b, c);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
  scanf("%d %d", &n, &k);
  set<int> st;
  for (int i = 0; i < n; ++i) {
    st.insert(k + i);
  }
  vector<tuple<int, int, int>> ans;
  int c = k + 3 * n - 1;
  for (int i = n - 1; i >= 0; i -= 2) {
    int b = k + n + i;
    auto it = st.lower_bound(c - b + 1);
    if (it == st.begin()) {
      puts("-1");
      return 0;
    }
    --it;
    int a = *it;
    st.erase(it);
    ans.emplace_back(a, b, c);
    --c;
  }
  for (int i = n - 2; i >= 0; i -= 2) {
    int b = k + n + i;
    auto it = st.lower_bound(c - b + 1);
    if (it == st.begin()) {
      puts("-1");
      return 0;
    }
    --it;
    int a = *it;
    st.erase(it);
    ans.emplace_back(a, b, c);
    --c;
  }
  for (auto it : ans) {
    int a, b, c;
    tie(a, b, c) = it;
    printf("%d %d %d\n", a, b, c);
  }
  return 0;
}

```