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> t[100001 * 2];
set<pair<int, int> > se;
int main() {
  ios_base::sync_with_stdio(0);
  istream::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int x[n + 1], y[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> x[i] >> y[i];
    t[x[i]].push_back(i);
    t[y[i] + 1].push_back(-i);
  }
  vector<int> g;
  g.clear();
  for (int i = 1; i <= 200000; i++) {
    for (auto nxt : t[i]) {
      if (nxt < 0) {
        if (se.find({-y[-nxt], -nxt}) != se.end()) se.erase({-y[-nxt], -nxt});
      } else
        se.insert({-y[nxt], nxt});
    }
    while (int(se.size()) > k) {
      g.push_back((*se.begin()).second);
      se.erase(se.begin());
    }
  }
  cout << g.size() << endl;
  for (auto nxt : g) {
    cout << nxt << " ";
  }
}