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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Seg {
  int l, r;
  int id;
  friend bool operator<(const Seg &a, const Seg &b) {
    if (a.r == b.r) {
      if (a.l == b.l) return a.id < b.id;
      return a.l < b.l;
    }
    return a.r < b.r;
  }
} p[200000 + 5];
set<Seg> Q;
bool cmp(Seg a, Seg b) { return a.l < b.l; }
vector<int> ans;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int ed = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].l, &p[i].r);
    p[i].id = i + 1;
    ed = max(p[i].r, ed);
  }
  sort(p, p + n, cmp);
  set<Seg>::iterator it;
  int cnt = 0;
  for (int i = 1; i <= ed; i++) {
    if (cnt >= n) break;
    if (!Q.empty()) {
      it = Q.begin();
      while (it != Q.end()) {
        Seg now = (*it);
        ++it;
        if (now.r < i)
          Q.erase(now);
        else
          break;
      }
    }
    while (cnt < n && p[cnt].l <= i) {
      if (p[cnt].r >= i) {
        Q.insert(p[cnt]);
      }
      cnt++;
    }
    while ((int)(Q.size()) > k) {
      it = Q.end();
      --it;
      ans.push_back((*it).id);
      Q.erase(it);
    }
  }
  int sz = ans.size();
  printf("%d\n", sz);
  for (int i = 0; i < sz; i++) {
    printf("%d", ans[i]);
    if (i == sz - 1)
      printf("\n");
    else
      printf(" ");
  }
  return 0;
}