#include <bits/stdc++.h>
using std::bitset;
using std::cin;
using std::cout;
using std::deque;
using std::endl;
using std::fill;
using std::lower_bound;
using std::pair;
using std::priority_queue;
using std::queue;
using std::set;
using std::sort;
using std::stack;
using std::string;
using std::upper_bound;
using std::vector;
const int maxn = 200000 + 10;
struct node {
  int id;
  int l;
  int r;
};
node a[maxn];
bool del[maxn];
std::list<int> ban_list;
struct cmp {
  bool operator()(const node lhs, const node rhs) const {
    return lhs.r < rhs.r;
  }
};
int main(int argc, char** argv) {
  if (argc == 2 && strcmp(argv[1], "-debug") == 0) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  int n, k;
  scanf("%d%d", &n, &k);
  int max = 0;
  for (int i = 0; i < n; i++) {
    a[i].id = i;
    scanf("%d%d", &a[i].l, &a[i].r);
    max = std::max(max, a[i].r);
  }
  std::multiset<node, cmp> s;
  sort(a, a + n, [](const node lhs, const node rhs) {
    return lhs.l == rhs.l ? lhs.r > rhs.r : lhs.l < rhs.l;
  });
  int now = 0;
  for (int i = 1; i <= max; i++) {
    auto it = s.begin();
    while ((int)s.size() > 0 && it->r < i) {
      it = s.erase(it);
    }
    while (now < n && a[now].l == i) {
      s.insert(a[now]);
      ++now;
    }
    if (s.size() == 0) continue;
    it = s.end();
    --it;
    while ((int)s.size() > k) {
      ban_list.push_back(it->id);
      s.erase(it);
      if ((int)s.size() <= k) break;
      it = s.end();
      --it;
    }
  }
  printf("%u\n", ban_list.size());
  for (int i : ban_list) printf("%d ", i + 1);
  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 std::bitset;
using std::cin;
using std::cout;
using std::deque;
using std::endl;
using std::fill;
using std::lower_bound;
using std::pair;
using std::priority_queue;
using std::queue;
using std::set;
using std::sort;
using std::stack;
using std::string;
using std::upper_bound;
using std::vector;
const int maxn = 200000 + 10;
struct node {
  int id;
  int l;
  int r;
};
node a[maxn];
bool del[maxn];
std::list<int> ban_list;
struct cmp {
  bool operator()(const node lhs, const node rhs) const {
    return lhs.r < rhs.r;
  }
};
int main(int argc, char** argv) {
  if (argc == 2 && strcmp(argv[1], "-debug") == 0) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  int n, k;
  scanf("%d%d", &n, &k);
  int max = 0;
  for (int i = 0; i < n; i++) {
    a[i].id = i;
    scanf("%d%d", &a[i].l, &a[i].r);
    max = std::max(max, a[i].r);
  }
  std::multiset<node, cmp> s;
  sort(a, a + n, [](const node lhs, const node rhs) {
    return lhs.l == rhs.l ? lhs.r > rhs.r : lhs.l < rhs.l;
  });
  int now = 0;
  for (int i = 1; i <= max; i++) {
    auto it = s.begin();
    while ((int)s.size() > 0 && it->r < i) {
      it = s.erase(it);
    }
    while (now < n && a[now].l == i) {
      s.insert(a[now]);
      ++now;
    }
    if (s.size() == 0) continue;
    it = s.end();
    --it;
    while ((int)s.size() > k) {
      ban_list.push_back(it->id);
      s.erase(it);
      if ((int)s.size() <= k) break;
      it = s.end();
      --it;
    }
  }
  printf("%u\n", ban_list.size());
  for (int i : ban_list) printf("%d ", i + 1);
  return 0;
}

```