#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<tuple<int, int, int> > segments(n);
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    segments[i] = {l, r, i};
  }
  sort(segments.begin(), segments.end());
  set<tuple<int, int, int> > active;
  int i = 0;
  vector<int> removed;
  while (i < n) {
    int curr_x = get<0>(segments[i]);
    while (i < n && get<0>(segments[i]) == curr_x) {
      active.insert(
          {get<1>(segments[i]), get<0>(segments[i]), get<2>(segments[i])});
      i++;
    }
    while (active.size() > 0 && get<0>(*(active.begin())) < curr_x)
      active.erase(active.begin());
    while (active.size() > k) {
      auto it = active.end();
      it--;
      removed.push_back(get<2>(*it));
      active.erase(it);
    }
    while (active.size() > 0 && get<0>(*(active.begin())) == curr_x)
      active.erase(active.begin());
  }
  cout << removed.size() << '\n';
  for (int i : removed) cout << i + 1 << ' ';
  cout << '\n';
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

The only difference between easy and hard versions is constraints.

You are given n segments on the coordinate axis OX. Segments can intersect, lie inside each other and even coincide. The i-th segment is [l_i; r_i] (l_i ≤ r_i) and it covers all integer points j such that l_i ≤ j ≤ r_i.

The integer point is called bad if it is covered by strictly more than k segments.

Your task is to remove the minimum number of segments so that there are no bad points at all.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 200) — the number of segments and the maximum number of segments by which each integer point can be covered.

The next n lines contain segments. The i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 200) — the endpoints of the i-th segment.

Output

In the first line print one integer m (0 ≤ m ≤ n) — the minimum number of segments you need to remove so that there are no bad points.

In the second line print m distinct integers p_1, p_2, ..., p_m (1 ≤ p_i ≤ n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.

Examples

Input


7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9


Output


3
1 4 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 2 4 


Input


6 1
2 3
3 3
2 3
2 2
2 3
2 3


Output


4
1 3 5 6 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<tuple<int, int, int> > segments(n);
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    segments[i] = {l, r, i};
  }
  sort(segments.begin(), segments.end());
  set<tuple<int, int, int> > active;
  int i = 0;
  vector<int> removed;
  while (i < n) {
    int curr_x = get<0>(segments[i]);
    while (i < n && get<0>(segments[i]) == curr_x) {
      active.insert(
          {get<1>(segments[i]), get<0>(segments[i]), get<2>(segments[i])});
      i++;
    }
    while (active.size() > 0 && get<0>(*(active.begin())) < curr_x)
      active.erase(active.begin());
    while (active.size() > k) {
      auto it = active.end();
      it--;
      removed.push_back(get<2>(*it));
      active.erase(it);
    }
    while (active.size() > 0 && get<0>(*(active.begin())) == curr_x)
      active.erase(active.begin());
  }
  cout << removed.size() << '\n';
  for (int i : removed) cout << i + 1 << ' ';
  cout << '\n';
  return 0;
}

```