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 long long mod = 1e9 + 7;
unsigned long long base = 131;
const int w = 2e5 + 5;
int n, m, k;
struct node {
  int x, y, z, t;
} s[w];
struct node2 {
  int x, y;
  bool friend operator<(node2 x, node2 y) { return x.x < y.x; }
};
int cmp(node x, node y) { return x.x < y.x; }
priority_queue<node2> qu;
int a[w];
int cmp3(node x, node y) { return x.t < y.t; }
int main() {
  int i, t, j;
  cin >> n >> m;
  int max1 = 0, min1 = 1e6;
  int r = 0;
  long long num = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &s[i].x, &s[i].y);
    s[i].z = 1;
    s[i].t = i;
    a[s[i].x]++, a[s[i].y + 1]--;
    max1 = max(max1, s[i].y);
    min1 = min(min1, s[i].x);
  }
  sort(s + 1, s + 1 + n, cmp);
  j = 1;
  for (i = min1; i <= max1; i++) {
    a[i] += a[i - 1];
    while (!qu.empty() && qu.top().x < i) qu.pop();
    while (j <= n && s[j].y >= i && s[j].x <= i) qu.push({s[j].y, j}), j++;
    if (a[i] > m) {
      while (a[i] > m) {
        node2 A = qu.top();
        qu.pop();
        a[i]--, a[A.x + 1]++;
        s[A.y].z = 0;
        num++;
      }
    }
  }
  cout << num << endl;
  sort(s + 1, s + 1 + n, cmp3);
  for (i = 1; i <= n; i++)
    if (s[i].z == 0) printf("%d%c", s[i].t, i == n ? '\n' : ' ');
  return 0;
}