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;
int n, k, a[200100], last = 1;
vector<int> ans;
vector<pair<int, int> > v[200100];
multiset<pair<int, int> > ms;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int L, R;
    cin >> L >> R;
    a[L]++;
    a[R + 1]--;
    v[L].push_back({R, i + 1});
  }
  for (int i = 1; i <= 2e5; i++) {
    a[i] += a[i - 1];
    for (pair<int, int> j : v[i]) ms.insert(j);
    while (a[i] > k) {
      pair<int, int> p = *(--ms.end());
      ms.erase(--ms.end());
      ans.push_back(p.second);
      a[i]--;
      a[p.first + 1]++;
    }
  }
  cout << ans.size() << endl;
  for (int x : ans) cout << x << " ";
  return 0;
}