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;
long long mod = 1000000007;
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
int bit(int x, int i) { return ((x >> (i - 1)) & 1); }
long long power(long long a, long long n, long long m) {
  long long ans = 1;
  while (n > 0) {
    if (n % 2) ans = (ans * a) % m;
    n = n >> 1;
    a = (a * a) % m;
  }
  return ans;
}
void merge(long long arr[], long long low, long long mid, long long high) {
  long long i = low, j = mid + 1, k = low;
  long long brr[100000];
  while (i <= mid && j <= high) {
    if (arr[i] <= arr[j])
      brr[k++] = arr[i++];
    else
      brr[k++] = arr[j++];
  }
  while (i <= mid) brr[k++] = arr[i++];
  while (j <= high) brr[k++] = arr[j++];
  for (k = low; k <= high; k++) arr[k] = brr[k];
}
void mergesort(long long arr[], long long low, long long high) {
  if (low < high) {
    long long mid = (low + high) / 2;
    mergesort(arr, low, mid);
    mergesort(arr, mid + 1, high);
    merge(arr, low, mid, high);
  }
}
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first < b.first;
}
int digit(int n) {
  int d = 0;
  while (n > 0) {
    d++;
    n = n / 10;
  }
  return d;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, l, r;
  cin >> n >> k;
  vector<pair<pair<long long, long long>, long long> > a;
  for (int i = 0; i < n; i++) {
    cin >> l >> r;
    a.push_back({{l, r}, i + 1});
  }
  sort(a.begin(), a.end());
  set<pair<long long, long long> > S;
  vector<int> res;
  for (auto it : a) {
    int left = it.first.first;
    int right = it.first.second;
    int idx = it.second;
    while (S.size() && S.begin()->first < left) S.erase(S.begin());
    S.insert({right, idx});
    if (S.size() > k) {
      auto i = S.end();
      i--;
      res.push_back(i->second);
      S.erase(i);
    }
  }
  cout << res.size() << "\n";
  for (auto i : res) cout << i << " ";
  return 0;
}