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;
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
char dir[] = {'U', 'R', 'D', 'L'};
const int N = 2e5 + 5;
int main() {
  file();
  fast();
  int n, k;
  cin >> n >> k;
  vector<vector<int>> in(N), out(N), a(n, vector<int>(2));
  for (int i = 0; i < n; i++) {
    cin >> a[i][0] >> a[i][1];
    in[a[i][0]].emplace_back(i);
    out[a[i][1] + 1].emplace_back(i);
  }
  vector<int> v, s(n);
  int sum = 0;
  priority_queue<pair<int, int>> q;
  for (int i = 1; i < N; i++) {
    for (int j : in[i]) {
      s[j] = 1;
      sum++;
      q.push({a[j][1] + 1, j});
    }
    for (int j : out[i]) {
      if (s[j]) sum--;
      s[j] = 0;
    }
    while (sum > k) {
      while (q.size() && s[q.top().second] == 0) q.pop();
      sum--;
      s[q.top().second] = 0;
      v.push_back(q.top().second);
      q.pop();
    }
  }
  cout << v.size() << "\n";
  sort((v).begin(), (v).end());
  for (int i : v) cout << i + 1 << " ";
}