#include <bits/stdc++.h>
using namespace std;
struct seg {
  int left, right;
  bool is_removed;
};
int main() {
  int n, k;
  cin >> n >> k;
  unordered_map<int, vector<int>> points_storage;
  unordered_map<int, seg> segments;
  vector<pair<int, int>> segs;
  vector<int> points;
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    segments.insert({i, {l, r}});
    segs.emplace_back(r, i);
  }
  sort(segs.rbegin(), segs.rend());
  for (auto segment : segs) {
    int index = segment.second;
    int l = segments[index].left, r = segments[index].right;
    for (int j = l; j <= r; j++) {
      if (points_storage.count(j)) {
        points_storage[j].push_back(index);
      } else {
        vector<int> s;
        s.push_back(index);
        points_storage.insert({j, s});
      }
    }
  }
  for (const auto& c : points_storage) {
    if (c.second.size() > k) {
      points.push_back(c.first);
    }
  }
  sort(points.begin(), points.end());
  vector<int> removed_segments;
  for (int p : points) {
    int coverage = points_storage[p].size();
    for (int j = coverage - 1; j >= 0; j--) {
      int curr_segment = points_storage[p][j];
      if (segments[curr_segment].is_removed) {
        coverage--;
      }
    }
    int i = 0;
    while (coverage > k) {
      int curr_segment = points_storage[p][i];
      if (!segments[curr_segment].is_removed) {
        segments[curr_segment].is_removed = true;
        removed_segments.push_back(curr_segment);
        coverage--;
      }
      i++;
    }
  }
  cout << removed_segments.size() << endl;
  for (auto removed : removed_segments) {
    cout << removed << " ";
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
struct seg {
  int left, right;
  bool is_removed;
};
int main() {
  int n, k;
  cin >> n >> k;
  unordered_map<int, vector<int>> points_storage;
  unordered_map<int, seg> segments;
  vector<pair<int, int>> segs;
  vector<int> points;
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    segments.insert({i, {l, r}});
    segs.emplace_back(r, i);
  }
  sort(segs.rbegin(), segs.rend());
  for (auto segment : segs) {
    int index = segment.second;
    int l = segments[index].left, r = segments[index].right;
    for (int j = l; j <= r; j++) {
      if (points_storage.count(j)) {
        points_storage[j].push_back(index);
      } else {
        vector<int> s;
        s.push_back(index);
        points_storage.insert({j, s});
      }
    }
  }
  for (const auto& c : points_storage) {
    if (c.second.size() > k) {
      points.push_back(c.first);
    }
  }
  sort(points.begin(), points.end());
  vector<int> removed_segments;
  for (int p : points) {
    int coverage = points_storage[p].size();
    for (int j = coverage - 1; j >= 0; j--) {
      int curr_segment = points_storage[p][j];
      if (segments[curr_segment].is_removed) {
        coverage--;
      }
    }
    int i = 0;
    while (coverage > k) {
      int curr_segment = points_storage[p][i];
      if (!segments[curr_segment].is_removed) {
        segments[curr_segment].is_removed = true;
        removed_segments.push_back(curr_segment);
        coverage--;
      }
      i++;
    }
  }
  cout << removed_segments.size() << endl;
  for (auto removed : removed_segments) {
    cout << removed << " ";
  }
  return 0;
}

```