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;
vector<int> st[200005], en[200005];
int l[200005], r[200005];
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  };
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    st[l[i]].push_back(i);
    en[r[i]].push_back(i);
  }
  set<pair<int, int> > second;
  vector<int> ans;
  for (int i = 1; i <= 200000; i++) {
    for (int j = 0; j < st[i].size(); j++) {
      int id = st[i][j];
      second.insert({-r[id], id});
    }
    while (second.size() > k) {
      auto it = second.begin();
      ans.push_back(it->second);
      second.erase({it->first, it->second});
    }
    for (int j = 0; j < en[i].size(); j++) {
      int id = en[i][j];
      second.erase({-r[id], id});
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
}