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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using ii = pair<int, int>;
using vii = vector<ii>;
struct Segment {
  int l, r, i;
  bool operator<(const Segment& o) const {
    if (l < o.l)
      return true;
    else if (l == o.l && r > o.r)
      return true;
    return false;
  }
};
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<Segment> a(n);
  for (int i = 0; i < (n); i++) cin >> a[i].l >> a[i].r, a[i].i = i + 1;
  sort(a.begin(), a.end());
  vi p(200002);
  for (int i = 0; i < (n); i++) p[a[i].l]++, p[a[i].r + 1]--;
  vi ans;
  map<int, vi> seg;
  int cur = 0, i = 0;
  for (int j = 0; j < ((int)p.size()); j++) {
    cur += p[j];
    while (i != n && a[i].l <= j) {
      seg[a[i].r + 1].push_back(a[i].i);
      i++;
    }
    while (cur > k) {
      auto take = seg.rbegin();
      p[take->first]++;
      ans.push_back(take->second.back());
      take->second.pop_back();
      if (take->second.empty()) seg.erase(take->first);
      cur--;
    }
  }
  cout << (int)ans.size() << '\n';
  for (int j : ans) cout << j << ' ';
  return 0;
}