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;
const int maxx = 800010;
int t, n, m, k, b, u;
int ans = 0, total = 0, cnt = 0;
int a[maxx], sum[maxx];
struct no {
  int num, l, r;
  bool operator<(const no &X) const { return r < X.r; }
};
priority_queue<no> q;
struct node {
  int num;
  int l;
  int r;
} s[maxx];
bool cmp(node a, node b) { return a.l < b.l; }
int main() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; i++) {
    scanf("%d%d", &s[i].l, &s[i].r);
    sum[s[i].l]++;
    sum[s[i].r + 1]--;
    s[i].num = i;
  }
  sort(s + 1, s + n + 1, cmp);
  cnt = 1;
  for (register int i = 1; i <= 200000; i++) {
    while (cnt <= n && s[cnt].l <= i) {
      q.push(no{s[cnt].num, s[cnt].l, s[cnt].r});
      cnt++;
    }
    sum[i] += sum[i - 1];
    while (sum[i] > k) {
      a[++total] = q.top().num;
      sum[i]--;
      sum[q.top().r + 1]++;
      q.pop();
    }
  }
  sort(a + 1, a + total + 1);
  printf("%d\n", total);
  for (register int i = 1; i <= total; i++) printf("%d ", a[i]);
  return 0;
}