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 kMaxN = 200001;
struct S {
  int i, l, r;
  bool d;
};
struct P {
  int i, v, t;
};
S s[kMaxN];
P p[kMaxN * 2];
int h[kMaxN];
int n, k, ans, t;
bool C(P p1, P p2) {
  if (p1.v != p2.v) {
    return p1.v < p2.v;
  };
  return p1.t > p2.t;
}
void Swap(int i, int j) {
  swap(h[i], h[j]);
  s[h[i]].i = i;
  s[h[j]].i = j;
}
void Down(int i) {
  int j;
  while ((j = (i << 1)) <= t) {
    if ((j < t) && (s[h[j]].r < s[h[j + 1]].r)) {
      j++;
    }
    if (s[h[i]].r >= s[h[j]].r) {
      break;
    }
    Swap(i, j);
    i = j;
  }
}
void Up(int i) {
  while ((i > 1) && (s[h[i / 2]].r < s[h[i]].r)) {
    Swap(i, i / 2);
    i /= 2;
  }
}
void Delete(int now) {
  int i = s[now].i;
  Swap(i, t);
  s[now].i = 0;
  t--;
  Up(i);
  Down(i);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> s[i].l >> s[i].r;
    p[i].i = p[i + n].i = i;
    p[i].t = 0;
    p[i + n].t = 1;
    p[i].v = s[i].l;
    p[i + n].v = s[i].r + 1;
  }
  sort(p + 1, p + 1 + 2 * n, C);
  t = ans = 0;
  for (int i = 1; i <= 2 * n; i++) {
    if (!p[i].t) {
      t++;
      h[t] = p[i].i;
      s[p[i].i].i = t;
      Up(t);
      if (t > k) {
        ans++;
        s[h[1]].d = 1;
        Delete(h[1]);
      }
    } else if (s[p[i].i].i) {
      Delete(p[i].i);
    }
  }
  cout << ans << endl;
  for (int i = 1; i <= n; i++) {
    if (s[i].d) {
      cout << i << " ";
    }
  }
  cout << endl;
  return 0;
}