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 segment {
  int x, y, idx;
  bool operator<(const segment ss) const { return x < ss.x; }
};
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<segment> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].x >> v[i].y;
    v[i].idx = i;
  }
  sort(v.begin(), v.end());
  set<pair<int, int>> st;
  vector<int> rem;
  for (int i = 0, j = 0; i < n; i = j) {
    while (j < n && v[i].x == v[j].x) {
      st.insert(make_pair(v[j].y, v[j].idx));
      j++;
    }
    while (!st.empty() && st.begin()->first < v[i].x) {
      st.erase(st.begin());
    }
    while ((int)st.size() > k) {
      rem.push_back(st.rbegin()->second);
      st.erase(*st.rbegin());
    }
  }
  cout << rem.size() << endl;
  for (auto i : rem) cout << i + 1 << ' ';
  return 0;
}