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 node {
  int l, r, id;
  bool operator<(const node &b) const { return r < b.r; }
} a[200009];
bool cmp(node a, node b) {
  if (a.l == b.l) return a.r < b.r;
  return a.l < b.l;
}
int vis[200009] = {}, ans[200009] = {};
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    vis[a[i].l]++;
    vis[a[i].r + 1]--;
    a[i].id = i;
  }
  sort(a, a + n, cmp);
  int j = 0, cnt = 0;
  priority_queue<node> q;
  for (int i = 1; i <= 200000; i++) {
    while (j < n && a[j].l <= i) q.push(a[j++]);
    vis[i] += vis[i - 1];
    while (vis[i] > k) {
      node p = q.top();
      q.pop();
      ans[cnt++] = p.id;
      vis[i]--;
      vis[p.r + 1]++;
    }
  }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++)
    printf("%d%s", ans[i] + 1, i == cnt - 1 ? "\n" : " ");
  return 0;
}