#include <bits/stdc++.h>
using namespace std;
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / GCD(a, b); }
int lb(int x) { return x & (-x); }
const long long N = 2e5 + 7;
const long long inf = 1e18 + 1;
const long long mod2 = 998244353;
const long long mod1 = 1e9 + 7;
const int P1 = 13331;
const int P2 = 131;
const double eps = 1e-6;
const double dbinf = 1e18 + 7;
const long long mod3 = 1e6 + 7;
int n, k;
int b[N];
vector<pair<int, int> > a[N];
priority_queue<pair<int, int> > q;
int main() {
  cin >> n >> k;
  for (int i = 1, l = 1, r = 1; i <= n; i++) {
    cin >> l >> r;
    b[l]++;
    b[r + 1]--;
    a[l].push_back({r, i});
  }
  vector<int> ans;
  for (int i = 1; i <= N - 1; i++) {
    b[i] += b[i - 1];
    while (!q.empty() && q.top().first < i) q.pop();
    for (auto& v : a[i]) q.push(v);
    while (b[i] > k) {
      pair<int, int> tmp = q.top();
      q.pop();
      ans.push_back(tmp.second);
      b[i]--;
      b[tmp.first + 1]++;
    }
  }
  cout << ans.size() << "\n";
  for (auto& v : ans) cout << v << " ";
  cout << "\n";
  return 0;
}

### 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;
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / GCD(a, b); }
int lb(int x) { return x & (-x); }
const long long N = 2e5 + 7;
const long long inf = 1e18 + 1;
const long long mod2 = 998244353;
const long long mod1 = 1e9 + 7;
const int P1 = 13331;
const int P2 = 131;
const double eps = 1e-6;
const double dbinf = 1e18 + 7;
const long long mod3 = 1e6 + 7;
int n, k;
int b[N];
vector<pair<int, int> > a[N];
priority_queue<pair<int, int> > q;
int main() {
  cin >> n >> k;
  for (int i = 1, l = 1, r = 1; i <= n; i++) {
    cin >> l >> r;
    b[l]++;
    b[r + 1]--;
    a[l].push_back({r, i});
  }
  vector<int> ans;
  for (int i = 1; i <= N - 1; i++) {
    b[i] += b[i - 1];
    while (!q.empty() && q.top().first < i) q.pop();
    for (auto& v : a[i]) q.push(v);
    while (b[i] > k) {
      pair<int, int> tmp = q.top();
      q.pop();
      ans.push_back(tmp.second);
      b[i]--;
      b[tmp.first + 1]++;
    }
  }
  cout << ans.size() << "\n";
  for (auto& v : ans) cout << v << " ";
  cout << "\n";
  return 0;
}

```