#include <bits/stdc++.h>
using namespace std;
int n, k;
struct node {
  int pos, x, y;
  node() {}
  node(int _x, int _y, int _pos) {
    x = _x;
    y = _y;
    pos = _pos;
  }
  bool operator<(const node &nd) const { return y < nd.y; }
};
bool cmp(node a, node b) {
  if (a.x < b.x) return true;
  if (a.x == b.x) return a.y > b.y;
  return false;
}
vector<node> v;
vector<int> ans;
bool used[int(2 * 1e5 + 5)];
int main() {
  int i, a, b, sz, tsz = 0;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a >> b;
    v.push_back(node(a, b, i));
    v.push_back(node(b, -1, i));
  }
  sort(v.begin(), v.end(), cmp);
  priority_queue<node> q;
  sz = v.size();
  for (i = 0; i < sz; i++) {
    node nd = v[i];
    if (nd.y != -1) {
      q.push(nd);
      tsz++;
      if (tsz > k) {
        ans.push_back(q.top().pos);
        used[q.top().pos] = true;
        q.pop();
        tsz--;
      }
    } else {
      if (!used[nd.pos]) tsz--;
    }
  }
  sz = ans.size();
  sort(ans.begin(), ans.end());
  cout << sz << endl;
  for (i = 0; i < sz; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int n, k;
struct node {
  int pos, x, y;
  node() {}
  node(int _x, int _y, int _pos) {
    x = _x;
    y = _y;
    pos = _pos;
  }
  bool operator<(const node &nd) const { return y < nd.y; }
};
bool cmp(node a, node b) {
  if (a.x < b.x) return true;
  if (a.x == b.x) return a.y > b.y;
  return false;
}
vector<node> v;
vector<int> ans;
bool used[int(2 * 1e5 + 5)];
int main() {
  int i, a, b, sz, tsz = 0;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a >> b;
    v.push_back(node(a, b, i));
    v.push_back(node(b, -1, i));
  }
  sort(v.begin(), v.end(), cmp);
  priority_queue<node> q;
  sz = v.size();
  for (i = 0; i < sz; i++) {
    node nd = v[i];
    if (nd.y != -1) {
      q.push(nd);
      tsz++;
      if (tsz > k) {
        ans.push_back(q.top().pos);
        used[q.top().pos] = true;
        q.pop();
        tsz--;
      }
    } else {
      if (!used[nd.pos]) tsz--;
    }
  }
  sz = ans.size();
  sort(ans.begin(), ans.end());
  cout << sz << endl;
  for (i = 0; i < sz; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}

```