#include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const long long int MOD = 1e9 + 7, INF = 1e18;
long long int dx[8] = {0, -1, 1, 0, 1, -1, 1, -1},
              dy[8] = {1, 0, 0, -1, -1, -1, 1, 1};
struct edge {
  long long int cost;
  long long int u, v;
};
const long long int MAX_SEG = 2e5;
long long int N, K, a, b;
vector<pair<long long int, long long int> > seg[MAX_SEG];
int main() {
  cin >> N >> K;
  for (long long int i = (0), i_end_ = (N); i < i_end_; i++) {
    cin >> a >> b;
    a--;
    b--;
    seg[a].push_back({b, i});
  }
  vector<long long int> ans;
  set<pair<long long int, long long int> > st;
  for (long long int i = (0), i_end_ = (MAX_SEG); i < i_end_; i++) {
    for (auto itr : seg[i]) {
      st.insert(itr);
    }
    while (((long long int)(st).size()) > K) {
      auto itr = st.rbegin();
      ans.push_back((*itr).second);
      st.erase(*itr);
    }
    while (((long long int)(st).size()) != 0) {
      auto itr = st.begin();
      if ((*itr).first != i) break;
      st.erase(*itr);
    }
  }
  cout << ((long long int)(ans).size()) << '\n';
  for (long long int i = (0), i_end_ = (((long long int)(ans).size()));
       i < i_end_; i++) {
    cout << ans[i] + 1 << " ";
  }
  cout << '\n';
}

### 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;
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const long long int MOD = 1e9 + 7, INF = 1e18;
long long int dx[8] = {0, -1, 1, 0, 1, -1, 1, -1},
              dy[8] = {1, 0, 0, -1, -1, -1, 1, 1};
struct edge {
  long long int cost;
  long long int u, v;
};
const long long int MAX_SEG = 2e5;
long long int N, K, a, b;
vector<pair<long long int, long long int> > seg[MAX_SEG];
int main() {
  cin >> N >> K;
  for (long long int i = (0), i_end_ = (N); i < i_end_; i++) {
    cin >> a >> b;
    a--;
    b--;
    seg[a].push_back({b, i});
  }
  vector<long long int> ans;
  set<pair<long long int, long long int> > st;
  for (long long int i = (0), i_end_ = (MAX_SEG); i < i_end_; i++) {
    for (auto itr : seg[i]) {
      st.insert(itr);
    }
    while (((long long int)(st).size()) > K) {
      auto itr = st.rbegin();
      ans.push_back((*itr).second);
      st.erase(*itr);
    }
    while (((long long int)(st).size()) != 0) {
      auto itr = st.begin();
      if ((*itr).first != i) break;
      st.erase(*itr);
    }
  }
  cout << ((long long int)(ans).size()) << '\n';
  for (long long int i = (0), i_end_ = (((long long int)(ans).size()));
       i < i_end_; i++) {
    cout << ans[i] + 1 << " ";
  }
  cout << '\n';
}

```