#include <bits/stdc++.h>
using namespace std;
const long long base = 1e9 + 7;
const long long MAX_N = 2e5 + 7;
long long dd2[MAX_N];
void Solves() {
  long long n, k;
  cin >> n >> k;
  long long l[n + 1];
  long long r[n + 1];
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> > >
      pq1;
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >,
                 greater<pair<long long, long long> > >
      pq2;
  vector<pair<long long, long long> > dd[MAX_N];
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    dd[l[i]].push_back(pair<long long, long long>(r[i], i));
  }
  int countt = 0;
  int sum = 0;
  vector<pair<long long, long long> > ans;
  for (int i = 1; i <= 2e5; i++) {
    for (auto j : dd[i]) {
      pq1.push(j);
      pq2.push(j);
      sum++;
    }
    while (!pq2.empty()) {
      pair<long long, long long> gg = pq2.top();
      if (gg.first < i) {
        pq2.pop();
        if (dd2[gg.second] == 0) countt++;
        dd2[gg.second]++;
      } else
        break;
    }
    while (sum - countt > k) {
      pair<long long, long long> gg = pq1.top();
      pq1.pop();
      if (dd2[gg.second] == 0) countt++;
      dd2[gg.second]++;
      ans.push_back(gg);
    }
  }
  cout << ans.size() << "\n";
  for (auto j : ans) cout << j.second << " ";
}
int main() { Solves(); }

### Prompt

Generate a CPP solution to the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long base = 1e9 + 7;
const long long MAX_N = 2e5 + 7;
long long dd2[MAX_N];
void Solves() {
  long long n, k;
  cin >> n >> k;
  long long l[n + 1];
  long long r[n + 1];
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> > >
      pq1;
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >,
                 greater<pair<long long, long long> > >
      pq2;
  vector<pair<long long, long long> > dd[MAX_N];
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    dd[l[i]].push_back(pair<long long, long long>(r[i], i));
  }
  int countt = 0;
  int sum = 0;
  vector<pair<long long, long long> > ans;
  for (int i = 1; i <= 2e5; i++) {
    for (auto j : dd[i]) {
      pq1.push(j);
      pq2.push(j);
      sum++;
    }
    while (!pq2.empty()) {
      pair<long long, long long> gg = pq2.top();
      if (gg.first < i) {
        pq2.pop();
        if (dd2[gg.second] == 0) countt++;
        dd2[gg.second]++;
      } else
        break;
    }
    while (sum - countt > k) {
      pair<long long, long long> gg = pq1.top();
      pq1.pop();
      if (dd2[gg.second] == 0) countt++;
      dd2[gg.second]++;
      ans.push_back(gg);
    }
  }
  cout << ans.size() << "\n";
  for (auto j : ans) cout << j.second << " ";
}
int main() { Solves(); }

```