#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3fll;
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int MAX = (int)2e5 + 10;
int n, k;
vector<pair<int, int>> v;
vector<pair<int, pair<int, int>>> seg;
int main() {
  ios::sync_with_stdio(false);
  ;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back({a, b});
    seg.push_back({a, {0, i}});
    seg.push_back({b, {1, i}});
  }
  sort(seg.begin(), seg.end());
  set<pair<int, int>> dentro;
  vector<bool> rem(n, false);
  int at = 0;
  for (int i = 0; i < MAX; i++) {
    while (at < seg.size() and seg[at].first == i and
           seg[at].second.first == 0) {
      dentro.insert({v[seg[at].second.second].second, seg[at].second.second});
      at++;
    }
    while (dentro.size() > k) {
      auto tira = dentro.end();
      tira--;
      rem[(*tira).second] = true;
      dentro.erase(tira);
    }
    while (at < seg.size() and seg[at].first == i and
           seg[at].second.first == 1) {
      if (!rem[seg[at].second.second])
        dentro.erase({seg[at].first, seg[at].second.second});
      at++;
    }
  }
  vector<int> resp;
  for (int i = 0; i < n; i++) {
    if (rem[i]) {
      resp.push_back(i);
    }
  }
  cout << resp.size() << '\n';
  for (int e : resp) {
    cout << e + 1 << " ";
  }
  cout << '\n';
  return 0;
}

### 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 int INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3fll;
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int MAX = (int)2e5 + 10;
int n, k;
vector<pair<int, int>> v;
vector<pair<int, pair<int, int>>> seg;
int main() {
  ios::sync_with_stdio(false);
  ;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back({a, b});
    seg.push_back({a, {0, i}});
    seg.push_back({b, {1, i}});
  }
  sort(seg.begin(), seg.end());
  set<pair<int, int>> dentro;
  vector<bool> rem(n, false);
  int at = 0;
  for (int i = 0; i < MAX; i++) {
    while (at < seg.size() and seg[at].first == i and
           seg[at].second.first == 0) {
      dentro.insert({v[seg[at].second.second].second, seg[at].second.second});
      at++;
    }
    while (dentro.size() > k) {
      auto tira = dentro.end();
      tira--;
      rem[(*tira).second] = true;
      dentro.erase(tira);
    }
    while (at < seg.size() and seg[at].first == i and
           seg[at].second.first == 1) {
      if (!rem[seg[at].second.second])
        dentro.erase({seg[at].first, seg[at].second.second});
      at++;
    }
  }
  vector<int> resp;
  for (int i = 0; i < n; i++) {
    if (rem[i]) {
      resp.push_back(i);
    }
  }
  cout << resp.size() << '\n';
  for (int e : resp) {
    cout << e + 1 << " ";
  }
  cout << '\n';
  return 0;
}

```