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>
#pragma GCC optimize(2)
using namespace std;
const double pi = acos(-1.0);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int inf = 0x7f7f7f7f;
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
int n, m, t;
int i, j, k;
set<pair<int, int> > s;
vector<int> ans;
struct Segment {
  int l, r;
  int id;
  const bool operator<(const Segment &b) const { return l < b.l; }
} a[N];
int main() {
  while (scanf("%d%d", &n, &k) == 2) {
    int maxx = -1, minn = inf;
    for (i = 1; i <= n; i++) {
      scanf("%d%d", &a[i].l, &a[i].r);
      minn = min(a[i].l, minn);
      maxx = max(a[i].r, maxx);
      a[i].id = i;
    }
    sort(a + 1, a + 1 + n);
    int cur = 1;
    for (int i = minn; i <= maxx; i++) {
      while (cur <= n && a[cur].l <= i) {
        s.insert(make_pair(a[cur].r, a[cur].id));
        ++cur;
      }
      while (s.size() && (*s.begin()).first < i) {
        s.erase(s.begin());
      }
      while (s.size() > k) {
        int id = s.rbegin()->second;
        s.erase(*s.rbegin());
        ans.push_back(id);
      }
    }
    printf("%d\n", ans.size());
    sort(ans.begin(), ans.end());
    for (auto it : ans) {
      printf("%d ", it);
    }
    puts("");
  }
  return 0;
}