#include <bits/stdc++.h>
using namespace std;
int n, k;
int l[200000], r[200000];
vector<int> solve() {
  vector<pair<int, int>> segments;
  for (int i = 0; i < n; ++i) {
    segments.push_back({l[i], -(i + 1)});
    segments.push_back({r[i], i + 1});
  }
  sort(segments.begin(), segments.end());
  vector<int> result;
  map<int, int> segment_map;
  priority_queue<pair<int, int>> pq;
  for (pair<int, int> &segment : segments) {
    int x = segment.first, index = segment.second;
    if (index < 0) {
      segment_map[-index] = x;
      pq.push({r[-index - 1], -index});
    } else {
      segment_map.erase(index);
    }
    while (segment_map.size() > k) {
      pair<int, int> s = pq.top();
      if (segment_map.count(s.second) > 0) {
        segment_map.erase(s.second);
        result.push_back(s.second);
      }
      pq.pop();
    }
  }
  return result;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &l[i], &r[i]);
  }
  vector<int> result = solve();
  printf("%d\n", result.size());
  for (int value : result) {
    printf("%d ", value);
  }
  printf("\n");
  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;
int n, k;
int l[200000], r[200000];
vector<int> solve() {
  vector<pair<int, int>> segments;
  for (int i = 0; i < n; ++i) {
    segments.push_back({l[i], -(i + 1)});
    segments.push_back({r[i], i + 1});
  }
  sort(segments.begin(), segments.end());
  vector<int> result;
  map<int, int> segment_map;
  priority_queue<pair<int, int>> pq;
  for (pair<int, int> &segment : segments) {
    int x = segment.first, index = segment.second;
    if (index < 0) {
      segment_map[-index] = x;
      pq.push({r[-index - 1], -index});
    } else {
      segment_map.erase(index);
    }
    while (segment_map.size() > k) {
      pair<int, int> s = pq.top();
      if (segment_map.count(s.second) > 0) {
        segment_map.erase(s.second);
        result.push_back(s.second);
      }
      pq.pop();
    }
  }
  return result;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &l[i], &r[i]);
  }
  vector<int> result = solve();
  printf("%d\n", result.size());
  for (int value : result) {
    printf("%d ", value);
  }
  printf("\n");
  return 0;
}

```