#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int n, k, l, r, freq[MAXN], BP[MAXN], bSeg[MAXN], cnt, ri[MAXN];
set<pair<int, int> > st;
vector<pair<int, pair<int, int> > > x[MAXN];
vector<pair<int, int> > seg;
int main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> l >> r;
    seg.push_back(make_pair(l, r));
    freq[l]++, freq[r + 1]--;
    ri[r]++;
  }
  for (int i = 1; i < MAXN; i++) {
    freq[i] += freq[i - 1];
    if (freq[i] > k) BP[i] = 1;
  }
  for (int i = 1; i < MAXN; i++) BP[i] += BP[i - 1];
  for (int i = 0; i < n; i++) {
    bSeg[i] = BP[seg[i].second] - BP[seg[i].first - 1];
    x[seg[i].first].push_back(
        make_pair(bSeg[i], make_pair(seg[i].second, i + 1)));
  }
  int ans = 0;
  vector<int> idxx;
  for (int i = 1; i < MAXN; i++) {
    for (int j = 0; j < x[i].size(); j++) {
      cnt++;
      st.insert(make_pair(x[i][j].second.first, x[i][j].second.second));
    }
    cnt -= ri[i - 1];
    while (st.size() > k && cnt > k && freq[i] > k) {
      pair<int, int> temp = *st.rbegin();
      if (temp.first < i) {
        st.erase(temp);
      } else {
        cnt--;
        ans++;
        idxx.push_back(temp.second);
        ri[temp.first]--;
        st.erase(temp);
        freq[i]--;
      }
    }
  }
  cout << ans << endl;
  for (int i = 0; i < idxx.size(); i++) {
    cout << idxx[i] << ' ';
  }
  cout << endl;
  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;
const int MAXN = 2e5 + 5;
int n, k, l, r, freq[MAXN], BP[MAXN], bSeg[MAXN], cnt, ri[MAXN];
set<pair<int, int> > st;
vector<pair<int, pair<int, int> > > x[MAXN];
vector<pair<int, int> > seg;
int main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> l >> r;
    seg.push_back(make_pair(l, r));
    freq[l]++, freq[r + 1]--;
    ri[r]++;
  }
  for (int i = 1; i < MAXN; i++) {
    freq[i] += freq[i - 1];
    if (freq[i] > k) BP[i] = 1;
  }
  for (int i = 1; i < MAXN; i++) BP[i] += BP[i - 1];
  for (int i = 0; i < n; i++) {
    bSeg[i] = BP[seg[i].second] - BP[seg[i].first - 1];
    x[seg[i].first].push_back(
        make_pair(bSeg[i], make_pair(seg[i].second, i + 1)));
  }
  int ans = 0;
  vector<int> idxx;
  for (int i = 1; i < MAXN; i++) {
    for (int j = 0; j < x[i].size(); j++) {
      cnt++;
      st.insert(make_pair(x[i][j].second.first, x[i][j].second.second));
    }
    cnt -= ri[i - 1];
    while (st.size() > k && cnt > k && freq[i] > k) {
      pair<int, int> temp = *st.rbegin();
      if (temp.first < i) {
        st.erase(temp);
      } else {
        cnt--;
        ans++;
        idxx.push_back(temp.second);
        ri[temp.first]--;
        st.erase(temp);
        freq[i]--;
      }
    }
  }
  cout << ans << endl;
  for (int i = 0; i < idxx.size(); i++) {
    cout << idxx[i] << ' ';
  }
  cout << endl;
  return 0;
}

```