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;
void blowback() {
  long long i, j, l, k, n, x, y, m, ans;
  string s;
  cin >> n >> m;
  pair<pair<long long, long long>, long long> a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i;
  }
  sort(a, a + n);
  multiset<pair<long long, long long> > ms;
  vector<long long> v;
  for (i = 1, j = 0; i <= 200000; i++) {
    while (a[j].first.first <= i) {
      ms.insert({a[j].first.second, a[j].second});
      j++;
    }
    while ((*ms.begin()).first < i) {
      ms.erase(ms.begin());
    }
    while (ms.size() > m) {
      auto it = ms.end();
      it--;
      v.push_back((*it).second);
      ms.erase(it);
    }
  }
  sort((v).begin(), (v).end());
  cout << v.size() << "\n";
  for (i = 0; i < v.size(); i++) cout << v[i] + 1 << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) {
    blowback();
  }
  return 0;
}