#include <bits/stdc++.h>
using namespace std;
int cnt[200005];
int num[200005];
int already[200005];
int main() {
  int n, k;
  cin >> n >> k;
  vector<vector<pair<int, int>>> vi;
  vector<vector<pair<int, int>>> vr;
  vi.resize(200005);
  vr.resize(200005);
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    cnt[l]++;
    cnt[r + 1]--;
    vi[l].push_back(make_pair(r, i));
    vr[r + 1].push_back(make_pair(r, i));
  }
  for (int i = 1; i < 200005; i++) {
    num[i] = num[i - 1] + cnt[i];
  }
  multiset<pair<int, int>> s;
  int removed = 0;
  int ans = 0;
  vector<int> finale;
  for (int i = 1; i < 200005; i++) {
    removed = removed - already[i];
    for (auto u : vi[i]) {
      s.insert(u);
    }
    for (auto u : vr[i]) {
      auto it = s.find(u);
      if (it != s.end()) {
        s.erase(it);
      }
    }
    while (num[i] - removed > k) {
      ans++;
      auto it = s.end();
      it--;
      int index = (*it).second;
      finale.push_back(index);
      removed++;
      already[(*it).first + 1]++;
      s.erase(it);
    }
  }
  cout << ans << endl;
  for (auto u : finale) {
    cout << u << " ";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 ≤ 2 ⋅ 10^5) — 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 ≤ 2 ⋅ 10^5) — 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
4 6 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 4 5 


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 cnt[200005];
int num[200005];
int already[200005];
int main() {
  int n, k;
  cin >> n >> k;
  vector<vector<pair<int, int>>> vi;
  vector<vector<pair<int, int>>> vr;
  vi.resize(200005);
  vr.resize(200005);
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    cnt[l]++;
    cnt[r + 1]--;
    vi[l].push_back(make_pair(r, i));
    vr[r + 1].push_back(make_pair(r, i));
  }
  for (int i = 1; i < 200005; i++) {
    num[i] = num[i - 1] + cnt[i];
  }
  multiset<pair<int, int>> s;
  int removed = 0;
  int ans = 0;
  vector<int> finale;
  for (int i = 1; i < 200005; i++) {
    removed = removed - already[i];
    for (auto u : vi[i]) {
      s.insert(u);
    }
    for (auto u : vr[i]) {
      auto it = s.find(u);
      if (it != s.end()) {
        s.erase(it);
      }
    }
    while (num[i] - removed > k) {
      ans++;
      auto it = s.end();
      it--;
      int index = (*it).second;
      finale.push_back(index);
      removed++;
      already[(*it).first + 1]++;
      s.erase(it);
    }
  }
  cout << ans << endl;
  for (auto u : finale) {
    cout << u << " ";
  }
  return 0;
}

```