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;
const int maxd = 2e5 + 10;
vector<pair<int, int> > a[maxd], b[maxd];
vector<int> ans;
set<pair<int, int> > s;
set<pair<int, int> >::iterator it;
int vis[maxd];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    a[x].push_back(pair<int, int>(y, i));
    b[y].push_back(pair<int, int>(y, i));
  }
  for (int i = 1; i <= 200000; i++) {
    for (pair<int, int> p : a[i]) s.insert(p);
    while (s.size() > m) {
      it = s.end();
      it--;
      vis[it->second] = 1;
      ans.push_back(it->second);
      s.erase(it);
    }
    for (pair<int, int> p : b[i])
      if (!vis[p.second]) s.erase(p);
  }
  printf("%lu\n", ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}