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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int64_t N = 200000;
vector<pair<int64_t, int64_t> > s[N + 5];
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int64_t n, k;
  cin >> n >> k;
  for (int64_t i = 0; i < n; i++) {
    int64_t l, r;
    cin >> l >> r;
    s[l].push_back({r + 1, i + 1});
  }
  set<pair<int64_t, int64_t> > b;
  vector<int64_t> rem;
  for (int64_t i = 1; i <= N; i++) {
    for (pair<int64_t, int64_t> x : s[i]) b.insert(x);
    while (true) {
      auto it = b.lower_bound({i, -1});
      if (it != b.end() && it->first == i)
        b.erase(it);
      else
        break;
    }
    while (b.size() > k) {
      rem.push_back(b.rbegin()->second);
      b.erase(*b.rbegin());
    }
  }
  cout << rem.size() << "\n";
  for (int64_t x : rem) cout << x << " ";
}