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;
template <class T>
void dbs(string str, T t) {
  cerr << str << " : " << t << "\n";
}
template <class T, class... S>
void dbs(string str, T t, S... s) {
  int idx = str.find(',');
  cerr << str.substr(0, idx) << " : " << t << ", ";
  dbs(str.substr(idx + 1), s...);
}
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class S, class T>
ostream& operator<<(ostream& os, const map<S, T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class T>
void prc(T a, T b) {
  cerr << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cerr << ", ";
    cerr << *i;
  }
  cerr << "]\n";
}
void code() {
  int i, j, k = 3, q = 1, n = 10, ans = 0, m, cur = 0;
  cin >> n >> k;
  std::vector<pair<int, int> > v;
  int l[n], r[n];
  for (i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    v.push_back({l[i], -(i + 1)});
    v.push_back({r[i], (i + 1)});
  }
  sort((v).begin(), (v).end());
  set<pair<int, int> > se;
  std::vector<int> remo;
  for (i = 0; i < 2 * n; i++) {
    int fir = v[i].first, sec = v[i].second;
    if (sec > 0) {
      if (((se).find(make_pair(r[sec - 1], sec - 1)) != (se).end())) {
        se.erase({r[sec - 1], sec - 1});
      }
    } else {
      sec *= -1;
      se.insert({r[sec - 1], sec - 1});
      if (se.size() > k) {
        auto xx = se.end();
        xx--;
        remo.push_back((xx->second) + 1);
        se.erase({r[xx->second], xx->second});
      }
    }
  }
  sort((remo).begin(), (remo).end());
  cout << remo.size() << "\n";
  for (i = 0; i < remo.size(); i++) cout << remo[i] << " ";
  return;
}
int main() {
  int tt, t = 1;
  for (tt = 1; tt <= t; tt++) {
    code();
  }
  return 0;
}