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;
template <class T>
bool mini(T& a, T b) {
  return a > b ? (a = b, 1) : 0;
}
template <class T>
bool maxi(T& a, T b) {
  return a < b ? (a = b, 1) : 0;
}
const int INF = 1e9 + 7;
const int N = 2e5 + 3;
int n, k;
vector<pair<int, int> > a[N];
vector<int> ans;
set<pair<int, int> > s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (1), _b = (n); i <= (_b); ++i) {
    int l, r;
    cin >> l >> r;
    a[l].push_back(pair<int, int>(r, i));
  }
  for (int i = (1), _b = (N); i < (_b); ++i) {
    while (!s.empty() && (s.begin()->first) < i) s.erase(s.begin());
    for (int j = (0), _b = ((int)a[i].size()); j < (_b); ++j) s.insert(a[i][j]);
    while ((int)s.size() > k) {
      pair<int, int> tmp = *(--s.end());
      s.erase(tmp);
      ans.push_back(tmp.second);
    }
  }
  sort(ans.begin(), ans.end());
  cout << (int)ans.size() << '\n';
  for (int i = (0), _b = ((int)ans.size()); i < (_b); ++i)
    cout << ans[i] << ' ';
  return 0;
}