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;
struct segm {
  int l = 0, r = 0, index = 0;
  bool operator<(const segm a) const {
    if (a.l == l) {
      return r < a.r;
    }
    return l < a.l;
  }
};
int n, k;
vector<segm> segms;
set<pair<int, int>> open;
vector<int> ans;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  segm p;
  for (int i = 1; i <= n; i++) {
    cin >> p.l >> p.r;
    p.index = i;
    segms.push_back(p);
  }
  sort(segms.begin(), segms.end());
  for (int i = 0; i < n; i++) {
    while (!open.empty()) {
      if (open.begin()->first >= segms[i].l) {
        break;
      }
      open.erase(open.begin());
    }
    open.insert({segms[i].r, segms[i].index});
    if (open.size() > k) {
      ans.push_back(prev(open.end())->second);
      open.erase(prev(open.end()));
    }
  }
  cout << ans.size() << endl;
  for (auto i : ans) {
    cout << i << " ";
  }
  cout << endl;
  return 0;
}