#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int dr8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dc8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dr4[] = {0, 1, -1, 0};
int dc4[] = {1, 0, 0, -1};
int pts[200010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  int l, r;
  int ans = 0;
  int mm = 200005;
  priority_queue<pair<int, int> > pq;
  memset(pts, 0, sizeof pts);
  priority_queue<int> pending;
  vector<vector<pair<int, int> > > all(mm);
  for (int i = 1; i <= n; ++i) {
    cin >> l >> r;
    all[l].push_back({r, i});
    ++pts[l];
    --pts[r + 1];
  }
  int csum = 0;
  for (int i = 1; i <= mm; ++i) {
    csum += pts[i];
    pts[i] = csum;
  }
  vector<int> inds;
  for (int i = 1; i <= mm; ++i) {
    for (auto ss : all[i]) {
      pq.push(ss);
    }
    while (pts[i] - (int)pending.size() > k) {
      ++ans;
      pending.push(-pq.top().first);
      inds.push_back(pq.top().second);
      pq.pop();
    }
    while (!pending.empty() and -pending.top() <= i) {
      pending.pop();
    }
  }
  cout << ans << '\n';
  for (auto ind : inds) {
    cout << ind << " ";
  }
  cout << "\n";
  return 0;
}

### Prompt

Your task is to create 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 int N = 200005;
int dr8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dc8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dr4[] = {0, 1, -1, 0};
int dc4[] = {1, 0, 0, -1};
int pts[200010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  int l, r;
  int ans = 0;
  int mm = 200005;
  priority_queue<pair<int, int> > pq;
  memset(pts, 0, sizeof pts);
  priority_queue<int> pending;
  vector<vector<pair<int, int> > > all(mm);
  for (int i = 1; i <= n; ++i) {
    cin >> l >> r;
    all[l].push_back({r, i});
    ++pts[l];
    --pts[r + 1];
  }
  int csum = 0;
  for (int i = 1; i <= mm; ++i) {
    csum += pts[i];
    pts[i] = csum;
  }
  vector<int> inds;
  for (int i = 1; i <= mm; ++i) {
    for (auto ss : all[i]) {
      pq.push(ss);
    }
    while (pts[i] - (int)pending.size() > k) {
      ++ans;
      pending.push(-pq.top().first);
      inds.push_back(pq.top().second);
      pq.pop();
    }
    while (!pending.empty() and -pending.top() <= i) {
      pending.pop();
    }
  }
  cout << ans << '\n';
  for (auto ind : inds) {
    cout << ind << " ";
  }
  cout << "\n";
  return 0;
}

```