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;
bool comp(pair<long long, pair<long long, long long>> a,
          pair<long long, pair<long long, long long>> b) {
  if (a.second.first != b.second.first) return a.second.first < b.second.first;
  return a.second.second < b.second.second;
}
void sol() {
  long long n, m, i, j, k;
  cin >> n >> k;
  vector<pair<long long, pair<long long, long long>>> v;
  for (i = 0; i < n; i++) {
    cin >> m >> j;
    v.push_back({i, {m, j}});
  }
  sort(v.begin(), v.end(), comp);
  set<pair<long long, long long>> q;
  vector<long long> ans;
  for (i = 0; i < n;) {
    while (q.size()) {
      pair<long long, long long> p = *q.begin();
      if (p.first < v[i].second.first)
        q.erase(p);
      else
        break;
    }
    long long x = v[i].second.first;
    q.insert({v[i].second.second, v[i].first});
    i++;
    while (q.size() > k) {
      pair<long long, long long> p = *q.rbegin();
      q.erase(p);
      ans.push_back(p.second);
    }
  }
  cout << ans.size() << "\n";
  for (auto i : ans) cout << i + 1 << " ";
}
int main() {
  sol();
  return 0;
}