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 fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int n, k;
int l[(int)2e5 + 100], r[(int)2e5 + 100], deleted[(int)2e5 + 100] = {0};
struct comper {
  bool operator()(const int& a, const int& b) const {
    if (r[a] == r[b]) return l[a] > l[b];
    return r[a] > r[b];
  }
};
int main() {
  cin >> n >> k;
  map<int, vector<int>> m1, m2;
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    m1[l[i]].push_back(i);
    m2[r[i] + 1].push_back(i);
  }
  set<pair<pair<int, int>, int>> s;
  for (int i = 1; i < (int)2e5 + 10; i++) {
    if (!m1[i].empty()) {
      for (int x : m1[i]) s.insert({{-r[x], l[x]}, x});
    }
    if (!m2[i].empty()) {
      for (int x : m2[i]) {
        s.erase({{-r[x], l[x]}, x});
      }
    }
    while ((int)((s).size()) > k) {
      int idx = (*s.begin()).second;
      deleted[idx] = 1;
      s.erase(s.begin());
    }
  }
  cout << accumulate(deleted, deleted + (int)2e5 + 100, 0) << endl;
  for (int i = 0; i < (int)2e5 + 2; i++)
    if (deleted[i]) cout << i + 1 << " ";
}