#include <bits/stdc++.h>
using namespace std;
typedef long double ld;
const long double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
long long mod = 1e9 + 7;
const long long inf = 1e18;
const long long maxn = 3e5 + 5;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, pair<long long, long long>>> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second.first;
    v[i].second.second = i + 1;
  }
  sort((v).begin(), (v).end());
  multiset<pair<long long, long long>> s;
  vector<long long> ans;
  for (auto i : v) {
    while (!s.empty() && (*s.begin()).first < i.first) s.erase(s.begin());
    while ((long long)((s).size()) > k) {
      auto it = --s.end();
      ans.push_back((*it).second);
      s.erase(it);
    }
    if ((long long)((s).size()) == k) {
      auto it = --s.end();
      if ((*it).first >= i.second.first) {
        ans.push_back((*it).second);
        s.erase(it);
        s.insert({i.second.first, i.second.second});
      } else
        ans.push_back(i.second.second);
    } else
      s.insert({i.second.first, i.second.second});
  }
  cout << (long long)((ans).size()) << "\n";
  for (auto i : ans) cout << i << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  while (tc--) {
    solve();
    cout << '\n';
  }
  return 0;
}

### 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;
typedef long double ld;
const long double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
long long mod = 1e9 + 7;
const long long inf = 1e18;
const long long maxn = 3e5 + 5;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, pair<long long, long long>>> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second.first;
    v[i].second.second = i + 1;
  }
  sort((v).begin(), (v).end());
  multiset<pair<long long, long long>> s;
  vector<long long> ans;
  for (auto i : v) {
    while (!s.empty() && (*s.begin()).first < i.first) s.erase(s.begin());
    while ((long long)((s).size()) > k) {
      auto it = --s.end();
      ans.push_back((*it).second);
      s.erase(it);
    }
    if ((long long)((s).size()) == k) {
      auto it = --s.end();
      if ((*it).first >= i.second.first) {
        ans.push_back((*it).second);
        s.erase(it);
        s.insert({i.second.first, i.second.second});
      } else
        ans.push_back(i.second.second);
    } else
      s.insert({i.second.first, i.second.second});
  }
  cout << (long long)((ans).size()) << "\n";
  for (auto i : ans) cout << i << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  while (tc--) {
    solve();
    cout << '\n';
  }
  return 0;
}

```