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 main() {
  long long int n, k;
  cin >> n >> k;
  vector<vector<long long int>> v;
  vector<pair<long long int, long long int>> seg;
  for (long long int i = 0; i < n; i++) {
    long long int a, b;
    cin >> a >> b;
    seg.push_back({a, b});
    v.push_back({a, 0, i});
    v.push_back({b, 1, i});
  }
  vector<long long int> ans;
  set<pair<long long int, long long int>> cur;
  sort(v.begin(), v.end());
  for (auto ar : v) {
    if (ar[1] == 0) {
      cur.insert({seg[ar[2]].second, ar[2]});
      if ((long long int)cur.size() > k) {
        auto j = cur.end();
        j--;
        ans.push_back(j->second);
        cur.erase(j);
      }
    } else {
      if (!cur.count({ar[0], ar[2]})) continue;
      cur.erase({ar[0], ar[2]});
    }
  }
  cout << ans.size() << "\n";
  for (auto i : ans) cout << i + 1 << " ";
  return 0;
}