#include <bits/stdc++.h>
using namespace std;
struct seg {
  int l, r, i;
  bool operator<(const seg& b) const { return l < b.l; }
};
int n, k;
seg a[200005];
bool u[400005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i].l >> a[i].r, a[i].i = i;
  sort(a, a + n);
  set<pair<int, int>> pq;
  for (int i = 0; i < k; i++) pq.insert({0, n + i});
  for (int i = 0; i < n; i++) {
    auto it = pq.lower_bound({a[i].l, -1});
    if (it == pq.begin()) {
      it = --pq.end();
      if (a[i].r < it->first) {
        u[it->second] = 1;
        pq.erase(it);
        pq.insert({a[i].r, a[i].i});
      } else {
        u[a[i].i] = 1;
      }
    } else {
      --it;
      pq.erase(it);
      pq.insert({a[i].r, a[i].i});
    }
  }
  cout << count(u, u + n, 1) << '\n';
  for (int i = 0; i < n; i++)
    if (u[i]) cout << i + 1 << ' ';
  cout << '\n';
}

### Prompt

Please formulate a Cpp solution 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;
struct seg {
  int l, r, i;
  bool operator<(const seg& b) const { return l < b.l; }
};
int n, k;
seg a[200005];
bool u[400005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i].l >> a[i].r, a[i].i = i;
  sort(a, a + n);
  set<pair<int, int>> pq;
  for (int i = 0; i < k; i++) pq.insert({0, n + i});
  for (int i = 0; i < n; i++) {
    auto it = pq.lower_bound({a[i].l, -1});
    if (it == pq.begin()) {
      it = --pq.end();
      if (a[i].r < it->first) {
        u[it->second] = 1;
        pq.erase(it);
        pq.insert({a[i].r, a[i].i});
      } else {
        u[a[i].i] = 1;
      }
    } else {
      --it;
      pq.erase(it);
      pq.insert({a[i].r, a[i].i});
    }
  }
  cout << count(u, u + n, 1) << '\n';
  for (int i = 0; i < n; i++)
    if (u[i]) cout << i + 1 << ' ';
  cout << '\n';
}

```