#include <bits/stdc++.h>
using namespace std;
long long n, k;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  vector<pair<pair<long long, long long>, long long>> v;
  multiset<long long> s1, s2;
  for (long long i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    v.push_back({{a, -b}, i + 1});
    s1.insert(a);
    s2.insert(b);
  }
  sort(v.begin(), v.end());
  vector<long long> ans;
  long long ct = 0;
  long long pos = 0;
  set<pair<long long, long long>> ss;
  while (s1.size() && s2.size()) {
    long long a = *s1.begin();
    long long b = *s2.begin();
    if (a <= b) {
      s1.erase(s1.find(a));
      ss.insert({v[pos].first.second, v[pos].second});
      if (ct == k) {
        ans.push_back((*ss.begin()).second);
        s2.erase(s2.find(-(*ss.begin()).first));
        ss.erase(ss.begin());
      } else
        ct++;
      pos++;
    } else {
      ct--;
      s2.erase(s2.find(b));
      if (ss.size() && (*ss.rbegin()).first == -b) ss.erase(*ss.rbegin());
    }
  }
  cout << ans.size() << '\n';
  for (long long i : ans) cout << i << " ";
}

### Prompt

Please create a solution in CPP 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;
long long n, k;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  vector<pair<pair<long long, long long>, long long>> v;
  multiset<long long> s1, s2;
  for (long long i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    v.push_back({{a, -b}, i + 1});
    s1.insert(a);
    s2.insert(b);
  }
  sort(v.begin(), v.end());
  vector<long long> ans;
  long long ct = 0;
  long long pos = 0;
  set<pair<long long, long long>> ss;
  while (s1.size() && s2.size()) {
    long long a = *s1.begin();
    long long b = *s2.begin();
    if (a <= b) {
      s1.erase(s1.find(a));
      ss.insert({v[pos].first.second, v[pos].second});
      if (ct == k) {
        ans.push_back((*ss.begin()).second);
        s2.erase(s2.find(-(*ss.begin()).first));
        ss.erase(ss.begin());
      } else
        ct++;
      pos++;
    } else {
      ct--;
      s2.erase(s2.find(b));
      if (ss.size() && (*ss.rbegin()).first == -b) ss.erase(*ss.rbegin());
    }
  }
  cout << ans.size() << '\n';
  for (long long i : ans) cout << i << " ";
}

```