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;
const int N = 2e5;
int n, m;
vector<pair<int, int> > v[N + 10];
set<pair<int, int> > s;
vector<int> ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    v[l].push_back(make_pair(r, i));
  }
  for (int i = 1; i <= N; i++) {
    for (int j = (int)v[i].size() - 1; j >= 0; j--) s.insert(v[i][j]);
    while (!s.empty() && s.begin()->first < i) s.erase(s.begin());
    while ((int)s.size() > m) {
      pair<int, int> c = *(--s.end());
      s.erase(c);
      ans.push_back(c.second);
    }
  }
  int sz = (int)ans.size();
  cout << sz << "\n";
  for (int i = 0; i < sz; i++) cout << ans[i] << " ";
  return 0;
}