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 PII = pair<int, int>;
const int N = 2e5 + 2;
vector<PII> seg[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int k, n;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    seg[l].emplace_back(r, i + 1);
  }
  set<PII> rs;
  vector<int> ans;
  for (int x = 1; x < N; x++) {
    for (auto& e : seg[x]) {
      rs.insert(e);
    }
    while (!rs.empty() && rs.begin()->first < x) {
      rs.erase(rs.begin());
    }
    while ((int)rs.size() > k) {
      auto e = *rs.rbegin();
      ans.push_back(e.second);
      rs.erase(e);
    }
  }
  cout << ans.size() << endl;
  for (int i : ans) cout << i << ' ';
  cout << endl;
  return 0;
}