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 ≤ 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct abc {
  bool lr;
  int coor, index, coor2;
};
bool comp(abc a, abc b) {
  if (a.coor != b.coor) return a.coor < b.coor;
  if (a.lr != b.lr) return a.lr < b.lr;
  if (a.coor2 != b.coor2) return a.coor2 > b.coor2;
  return a.index < b.index;
}
signed main() {
  int n, k;
  cin >> n >> k;
  long long int ans = 0;
  vector<abc> inn;
  for (int i = 1; i <= n; ++i) {
    int x, y;
    cin >> x >> y;
    abc h = {false, x, i, y};
    inn.push_back(h);
    h = {true, y, i, 0};
    inn.push_back(h);
  }
  sort(inn.begin(), inn.end(), comp);
  int coun = 0;
  set<int> del_ind, otkr;
  set<pair<int, int>> coor_ind;
  for (auto it : inn) {
    if (it.lr == false) {
      coun++;
      otkr.insert(it.index);
      coor_ind.insert(make_pair(it.coor2, it.index));
      if (coun > k) {
        coun--;
        ans++;
        auto it_ = coor_ind.end();
        it_--;
        otkr.erase(otkr.find((*it_).second));
        del_ind.insert((*it_).second);
        coor_ind.erase(it_);
      }
    } else {
      if (del_ind.find(it.index) == del_ind.end()) {
        coun--;
        otkr.erase(otkr.find(it.index));
        coor_ind.erase(coor_ind.find(make_pair(it.coor, it.index)));
      }
    }
  }
  cout << ans << "\n";
  for (auto it : del_ind) {
    cout << it << " ";
  }
}