#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, l, r[200005], cnt = 0;
  bool valid[200005];
  pair<int, int> p;
  pair<int, pair<int, int> > u;
  vector<int> v;
  vector<pair<int, pair<int, int> > > event;
  priority_queue<pair<int, int> > pq;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &l, &r[i]);
    event.push_back(make_pair(l, make_pair(0, i)));
    event.push_back(make_pair(r[i] + 1, make_pair(1, i)));
  }
  sort(event.begin(), event.end());
  memset(valid, true, sizeof valid);
  for (int i = 0; i < (int)event.size(); i++) {
    u = event[i];
    l = u.first;
    while (u.first == l) {
      if (u.second.first == 0) {
        pq.push(make_pair(r[u.second.second], u.second.second));
        cnt++;
      } else if (valid[u.second.second]) {
        valid[u.second.second] = false;
        cnt--;
      }
      i++;
      if (i == (int)event.size()) break;
      u = event[i];
    }
    i--;
    while (cnt > k) {
      p = pq.top();
      pq.pop();
      while (!valid[p.second]) {
        p = pq.top();
        pq.pop();
      }
      valid[p.second] = false;
      v.push_back(p.second);
      cnt--;
    }
  }
  printf("%d\n", (int)v.size());
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) putchar(' ');
    printf("%d", v[i]);
  }
  puts("");
  return 0;
}

### Prompt

Please formulate a cpp solution 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() {
  int n, k, l, r[200005], cnt = 0;
  bool valid[200005];
  pair<int, int> p;
  pair<int, pair<int, int> > u;
  vector<int> v;
  vector<pair<int, pair<int, int> > > event;
  priority_queue<pair<int, int> > pq;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &l, &r[i]);
    event.push_back(make_pair(l, make_pair(0, i)));
    event.push_back(make_pair(r[i] + 1, make_pair(1, i)));
  }
  sort(event.begin(), event.end());
  memset(valid, true, sizeof valid);
  for (int i = 0; i < (int)event.size(); i++) {
    u = event[i];
    l = u.first;
    while (u.first == l) {
      if (u.second.first == 0) {
        pq.push(make_pair(r[u.second.second], u.second.second));
        cnt++;
      } else if (valid[u.second.second]) {
        valid[u.second.second] = false;
        cnt--;
      }
      i++;
      if (i == (int)event.size()) break;
      u = event[i];
    }
    i--;
    while (cnt > k) {
      p = pq.top();
      pq.pop();
      while (!valid[p.second]) {
        p = pq.top();
        pq.pop();
      }
      valid[p.second] = false;
      v.push_back(p.second);
      cnt--;
    }
  }
  printf("%d\n", (int)v.size());
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) putchar(' ');
    printf("%d", v[i]);
  }
  puts("");
  return 0;
}

```