#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int l[N], r[N], d[N];
priority_queue<pair<int, int> > pq;
vector<pair<int, int> > G[N];
vector<int> ans;
int main() {
  int n, k, mx = 0;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    G[l[i]].push_back(pair<int, int>(r[i], i + 1));
    d[l[i]]++, d[r[i] + 1]--;
    mx = max(mx, r[i]);
  }
  for (int i = 1; i <= mx; i++) {
    d[i] += d[i - 1];
    for (pair<int, int> u : G[i]) pq.push(u);
    if (d[i] > k) {
      int over = d[i] - k;
      while (over--) {
        pair<int, int> temp = pq.top();
        ans.push_back(temp.second);
        d[i]--;
        d[temp.first + 1]++;
        pq.pop();
      }
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d%c", ans[i], i == ans.size() - 1 ? '\n' : ' ');
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = 2e5 + 5;
int l[N], r[N], d[N];
priority_queue<pair<int, int> > pq;
vector<pair<int, int> > G[N];
vector<int> ans;
int main() {
  int n, k, mx = 0;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    G[l[i]].push_back(pair<int, int>(r[i], i + 1));
    d[l[i]]++, d[r[i] + 1]--;
    mx = max(mx, r[i]);
  }
  for (int i = 1; i <= mx; i++) {
    d[i] += d[i - 1];
    for (pair<int, int> u : G[i]) pq.push(u);
    if (d[i] > k) {
      int over = d[i] - k;
      while (over--) {
        pair<int, int> temp = pq.top();
        ans.push_back(temp.second);
        d[i]--;
        d[temp.first + 1]++;
        pq.pop();
      }
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d%c", ans[i], i == ans.size() - 1 ? '\n' : ' ');
  }
  return 0;
}

```