#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 2;
vector<pair<int, int> > segs[N];
void print(set<pair<int, int> > rs) {
  set<pair<int, int> >::iterator ss;
  for (ss = rs.begin(); ss != rs.end(); ss++) {
    cout << ss->first << " ";
  }
  cout << endl;
}
void solve() {
  int n, k;
  cin >> n >> k;
  int minn = INT_MAX, maxx = INT_MIN;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    if (l < minn) minn = l;
    if (r > maxx) maxx = r;
    segs[l].push_back(make_pair(r, i + 1));
  }
  set<pair<int, int> > rs;
  vector<int> ans;
  for (int i = minn; i <= maxx; i++) {
    vector<pair<int, int> >::iterator itr;
    for (itr = segs[i].begin(); itr != segs[i].end(); itr++) {
      rs.insert(*itr);
    }
    while (rs.size() != 0 && rs.begin()->first < i) {
      rs.erase(rs.begin());
    }
    while (rs.size() > k) {
      set<pair<int, int> >::iterator ss = --rs.end();
      ans.push_back(ss->second);
      rs.erase(ss);
    }
  }
  cout << ans.size() << endl;
  vector<int>::iterator itr;
  for (itr = ans.begin(); itr != ans.end(); itr++) {
    cout << *itr << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 N = 2e5 + 2;
vector<pair<int, int> > segs[N];
void print(set<pair<int, int> > rs) {
  set<pair<int, int> >::iterator ss;
  for (ss = rs.begin(); ss != rs.end(); ss++) {
    cout << ss->first << " ";
  }
  cout << endl;
}
void solve() {
  int n, k;
  cin >> n >> k;
  int minn = INT_MAX, maxx = INT_MIN;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    if (l < minn) minn = l;
    if (r > maxx) maxx = r;
    segs[l].push_back(make_pair(r, i + 1));
  }
  set<pair<int, int> > rs;
  vector<int> ans;
  for (int i = minn; i <= maxx; i++) {
    vector<pair<int, int> >::iterator itr;
    for (itr = segs[i].begin(); itr != segs[i].end(); itr++) {
      rs.insert(*itr);
    }
    while (rs.size() != 0 && rs.begin()->first < i) {
      rs.erase(rs.begin());
    }
    while (rs.size() > k) {
      set<pair<int, int> >::iterator ss = --rs.end();
      ans.push_back(ss->second);
      rs.erase(ss);
    }
  }
  cout << ans.size() << endl;
  vector<int>::iterator itr;
  for (itr = ans.begin(); itr != ans.end(); itr++) {
    cout << *itr << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

```