#include <bits/stdc++.h>
using namespace std;
template <class T>
bool mini(T& a, T b) {
  return a > b ? (a = b, 1) : 0;
}
template <class T>
bool maxi(T& a, T b) {
  return a < b ? (a = b, 1) : 0;
}
const int INF = 1e9 + 7;
int n, k;
vector<pair<pair<int, int>, int> > a;
vector<int> ans;
multiset<int> s;
bool cmp(const pair<pair<int, int>, int>& a,
         const pair<pair<int, int>, int>& b) {
  return (a.first.second < b.first.second) ||
         (a.first.second == b.first.second && a.first.first < b.first.first);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (0), _b = (n); i < (_b); ++i) {
    int l, r;
    cin >> l >> r;
    a.push_back(pair<pair<int, int>, int>(pair<int, int>(l, r), i));
  }
  sort(a.begin(), a.end(), cmp);
  for (int i = (0), _b = (k); i < (_b); ++i) s.insert(0);
  vector<bool> vis(n);
  for (int i = (0), _b = (n); i < (_b); ++i) {
    multiset<int>::iterator it = s.lower_bound(a[i].first.first);
    if (it != s.begin()) {
      --it;
      s.erase(it);
      s.insert(a[i].first.second);
      vis[a[i].second] = 1;
    }
  }
  for (int i = (0), _b = (n); i < (_b); ++i)
    if (!vis[i]) ans.push_back(i);
  cout << (int)ans.size() << '\n';
  for (int i = (0), _b = ((int)ans.size()); i < (_b); ++i)
    cout << ans[i] + 1 << ' ';
  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;
template <class T>
bool mini(T& a, T b) {
  return a > b ? (a = b, 1) : 0;
}
template <class T>
bool maxi(T& a, T b) {
  return a < b ? (a = b, 1) : 0;
}
const int INF = 1e9 + 7;
int n, k;
vector<pair<pair<int, int>, int> > a;
vector<int> ans;
multiset<int> s;
bool cmp(const pair<pair<int, int>, int>& a,
         const pair<pair<int, int>, int>& b) {
  return (a.first.second < b.first.second) ||
         (a.first.second == b.first.second && a.first.first < b.first.first);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (0), _b = (n); i < (_b); ++i) {
    int l, r;
    cin >> l >> r;
    a.push_back(pair<pair<int, int>, int>(pair<int, int>(l, r), i));
  }
  sort(a.begin(), a.end(), cmp);
  for (int i = (0), _b = (k); i < (_b); ++i) s.insert(0);
  vector<bool> vis(n);
  for (int i = (0), _b = (n); i < (_b); ++i) {
    multiset<int>::iterator it = s.lower_bound(a[i].first.first);
    if (it != s.begin()) {
      --it;
      s.erase(it);
      s.insert(a[i].first.second);
      vis[a[i].second] = 1;
    }
  }
  for (int i = (0), _b = (n); i < (_b); ++i)
    if (!vis[i]) ans.push_back(i);
  cout << (int)ans.size() << '\n';
  for (int i = (0), _b = ((int)ans.size()); i < (_b); ++i)
    cout << ans[i] + 1 << ' ';
  return 0;
}

```