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 inf = 2e5 + 9;
int n, k;
vector<int> ans;
pair<pair<int, int>, int> a[inf];
set<pair<int, int> > add[inf], cur, del[inf];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    a[i] = make_pair(make_pair(x, y), i);
    add[x].insert(make_pair(y, i));
    del[y].insert(make_pair(y, i));
  }
  sort(a + 1, a + 1 + n);
  int cnt = 0;
  for (int i = 1; i < inf; i++) {
    cnt += add[i].size();
    for (auto o : add[i]) cur.insert(o);
    while (cnt > k) {
      auto o = *cur.rbegin();
      cur.erase(o);
      cnt--;
      ans.push_back(o.second);
      del[o.first].erase(o);
    }
    cnt -= del[i].size();
    for (auto o : del[i]) cur.erase(o);
  }
  cout << ans.size() << endl;
  for (auto o : ans) cout << o << " ";
  cout << endl;
}