#include <bits/stdc++.h>
using namespace std;
int t, n, k, l, r;
int L[200000 + 5];
int R[200000 + 5];
vector<int> st[200000 + 5];
vector<int> ed[200000 + 5];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n + 1; i++) {
    scanf("%d%d", &l, &r);
    L[i] = l;
    R[i] = r;
    st[l].push_back(i);
    ed[r].push_back(i);
  }
  set<pair<int, int> > cur;
  set<pair<int, int> >::iterator it;
  vector<int> ans;
  for (int i = 1; i < 200000 + 1; i++) {
    for (int j = 0; j < st[i].size(); j++) cur.insert({R[st[i][j]], st[i][j]});
    while (cur.size() > k) {
      it = cur.end();
      it--;
      pair<int, int> X = *it;
      ans.push_back(X.second);
      cur.erase(it);
    }
    for (int j = 0; j < ed[i].size(); j++) cur.erase({R[ed[i][j]], ed[i][j]});
  }
  printf("%d", ans.size());
  printf("\n");
  for (int i = 0; i < ans.size(); i++) {
    printf("%d", ans[i]);
    printf(" ");
  }
  printf("\n");
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 t, n, k, l, r;
int L[200000 + 5];
int R[200000 + 5];
vector<int> st[200000 + 5];
vector<int> ed[200000 + 5];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n + 1; i++) {
    scanf("%d%d", &l, &r);
    L[i] = l;
    R[i] = r;
    st[l].push_back(i);
    ed[r].push_back(i);
  }
  set<pair<int, int> > cur;
  set<pair<int, int> >::iterator it;
  vector<int> ans;
  for (int i = 1; i < 200000 + 1; i++) {
    for (int j = 0; j < st[i].size(); j++) cur.insert({R[st[i][j]], st[i][j]});
    while (cur.size() > k) {
      it = cur.end();
      it--;
      pair<int, int> X = *it;
      ans.push_back(X.second);
      cur.erase(it);
    }
    for (int j = 0; j < ed[i].size(); j++) cur.erase({R[ed[i][j]], ed[i][j]});
  }
  printf("%d", ans.size());
  printf("\n");
  for (int i = 0; i < ans.size(); i++) {
    printf("%d", ans[i]);
    printf(" ");
  }
  printf("\n");
  return 0;
}

```