#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const long long INF = 3e11 + 7;
const int base = 2e5 + 1;
const int MAX = 1e5 + 2;
const double EPS = 1e-9;
const double PI = acos(-1.);
const int MAXN = 2 * 1e5 + 5;
vector<vector<int> > l(MAXN), r(MAXN);
vector<int> res;
vector<pair<int, int> > v(MAXN);
bool was[MAXN];
struct by_r {
  bool operator()(int x, int y) {
    if (v[x].second == v[y].second) return v[x].first < v[y].first;
    return v[x].second > v[y].second;
  }
};
int main() {
  int n, k;
  cin >> n >> k;
  set<int> points;
  for (long long i = (0); i < (n); ++i) {
    cin >> v[i].first >> v[i].second;
    l[v[i].first].push_back(i);
    r[v[i].second + 1].push_back(i);
    points.insert(v[i].first);
    points.insert(v[i].second + 1);
  }
  set<pair<int, int> > cur;
  for (auto i : points) {
    for (auto to : l[i]) cur.insert({v[to].second, to});
    for (auto to : r[i]) cur.erase({v[to].second, to});
    while ((int)cur.size() > k) {
      pair<int, int> x = *(--cur.end());
      auto it = cur.end();
      it--;
      cur.erase(it);
      res.push_back(x.second);
    }
  }
  cout << (int)res.size() << endl;
  for (long long i = (0); i < ((int)res.size()); ++i) cout << res[i] + 1 << ' ';
  return 0;
}

### 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 MOD = 998244353;
const long long INF = 3e11 + 7;
const int base = 2e5 + 1;
const int MAX = 1e5 + 2;
const double EPS = 1e-9;
const double PI = acos(-1.);
const int MAXN = 2 * 1e5 + 5;
vector<vector<int> > l(MAXN), r(MAXN);
vector<int> res;
vector<pair<int, int> > v(MAXN);
bool was[MAXN];
struct by_r {
  bool operator()(int x, int y) {
    if (v[x].second == v[y].second) return v[x].first < v[y].first;
    return v[x].second > v[y].second;
  }
};
int main() {
  int n, k;
  cin >> n >> k;
  set<int> points;
  for (long long i = (0); i < (n); ++i) {
    cin >> v[i].first >> v[i].second;
    l[v[i].first].push_back(i);
    r[v[i].second + 1].push_back(i);
    points.insert(v[i].first);
    points.insert(v[i].second + 1);
  }
  set<pair<int, int> > cur;
  for (auto i : points) {
    for (auto to : l[i]) cur.insert({v[to].second, to});
    for (auto to : r[i]) cur.erase({v[to].second, to});
    while ((int)cur.size() > k) {
      pair<int, int> x = *(--cur.end());
      auto it = cur.end();
      it--;
      cur.erase(it);
      res.push_back(x.second);
    }
  }
  cout << (int)res.size() << endl;
  for (long long i = (0); i < ((int)res.size()); ++i) cout << res[i] + 1 << ' ';
  return 0;
}

```