#include <bits/stdc++.h>
using namespace std;
int l[200005], r[200005];
set<pair<int, int> > X[200005];
vector<int> ans;
int main() {
  int N, K, i, j, k;
  cin >> N >> K;
  for (i = 1; i <= N; i++) {
    cin >> l[i] >> r[i];
    X[l[i]].insert({r[i], i});
    X[r[i] + 1].insert({-1, i});
  }
  int cnt = 0;
  set<pair<int, int> > lol;
  for (i = 1; i <= 200000; i++) {
    if ((int)X[i].size()) {
      for (auto s : X[i]) {
        if (s.first < 0) --cnt;
      }
    }
    if ((int)X[i].size()) {
      for (auto s : X[i]) {
        if (s.first > 0) {
          ++cnt;
          lol.insert({s.first, s.second});
        }
      }
    }
    while (cnt > K) {
      cnt--;
      auto it = *lol.rbegin();
      ans.push_back(it.second);
      lol.erase({it.first, it.second});
      X[it.first + 1].erase({-1, it.second});
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  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;
int l[200005], r[200005];
set<pair<int, int> > X[200005];
vector<int> ans;
int main() {
  int N, K, i, j, k;
  cin >> N >> K;
  for (i = 1; i <= N; i++) {
    cin >> l[i] >> r[i];
    X[l[i]].insert({r[i], i});
    X[r[i] + 1].insert({-1, i});
  }
  int cnt = 0;
  set<pair<int, int> > lol;
  for (i = 1; i <= 200000; i++) {
    if ((int)X[i].size()) {
      for (auto s : X[i]) {
        if (s.first < 0) --cnt;
      }
    }
    if ((int)X[i].size()) {
      for (auto s : X[i]) {
        if (s.first > 0) {
          ++cnt;
          lol.insert({s.first, s.second});
        }
      }
    }
    while (cnt > K) {
      cnt--;
      auto it = *lol.rbegin();
      ans.push_back(it.second);
      lol.erase({it.first, it.second});
      X[it.first + 1].erase({-1, it.second});
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}

```