#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int st[N], ed[N];
struct Event {
  int x, k, id;
  Event(int x, int k, int id) : x(x), k(k), id(id) {}
  bool operator<(const Event &e) const {
    if (x != e.x) return x < e.x;
    if (k != e.k) return k < e.k;
    return id < e.id;
  }
};
set<Event> events;
inline void addEvent(int x, int k, int id) { events.emplace(x, k, id); }
inline void addSegment(int x0, int x1, int id) {
  addEvent(x0, 0, id);
  addEvent(x1, 1, id);
  st[id] = x0;
  ed[id] = x1;
}
vector<int> sweep(int mx) {
  auto ord = [&](const int &a, const int &b) {
    if (ed[a] != ed[b]) return ed[a] > ed[b];
    if (st[a] != st[b]) return st[a] < st[b];
    return a < b;
  };
  set<int, decltype(ord)> s(ord);
  vector<int> ans = {};
  for (Event e : events) {
    if (e.k == 0) s.emplace(e.id);
    while ((int)s.size() > mx) {
      ans.emplace_back(*s.begin());
      s.erase(s.begin());
    }
    auto it = s.lower_bound(e.id);
    if (e.k == 1 && it != s.end() && *it == e.id) s.erase(e.id);
  }
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    addSegment(l, r, i);
  }
  vector<int> ans = sweep(k);
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < (int)ans.size(); ++i)
    printf("%d%c", ans[i], " \n"[i + 1 == (int)ans.size()]);
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const int N = 2e5 + 7;
int st[N], ed[N];
struct Event {
  int x, k, id;
  Event(int x, int k, int id) : x(x), k(k), id(id) {}
  bool operator<(const Event &e) const {
    if (x != e.x) return x < e.x;
    if (k != e.k) return k < e.k;
    return id < e.id;
  }
};
set<Event> events;
inline void addEvent(int x, int k, int id) { events.emplace(x, k, id); }
inline void addSegment(int x0, int x1, int id) {
  addEvent(x0, 0, id);
  addEvent(x1, 1, id);
  st[id] = x0;
  ed[id] = x1;
}
vector<int> sweep(int mx) {
  auto ord = [&](const int &a, const int &b) {
    if (ed[a] != ed[b]) return ed[a] > ed[b];
    if (st[a] != st[b]) return st[a] < st[b];
    return a < b;
  };
  set<int, decltype(ord)> s(ord);
  vector<int> ans = {};
  for (Event e : events) {
    if (e.k == 0) s.emplace(e.id);
    while ((int)s.size() > mx) {
      ans.emplace_back(*s.begin());
      s.erase(s.begin());
    }
    auto it = s.lower_bound(e.id);
    if (e.k == 1 && it != s.end() && *it == e.id) s.erase(e.id);
  }
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    addSegment(l, r, i);
  }
  vector<int> ans = sweep(k);
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < (int)ans.size(); ++i)
    printf("%d%c", ans[i], " \n"[i + 1 == (int)ans.size()]);
  return 0;
}

```