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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = (int)1e9 + 7;
const int N = (int)2e5 + 7;
int n, k;
vector<pair<int, int> > vl[N];
int vr[N];
void Init() {
  memset(vr, 0, sizeof(vr));
  for (int i = (1); i < (N); ++i) vl[i].clear();
  for (int i = (0); i < (n); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    vl[l].push_back(make_pair(r, i + 1));
    vr[r + 1]++;
  }
}
int Solve() {
  vector<int> ans;
  priority_queue<pair<int, int> > Q;
  int cur = 0;
  for (int i = (1); i < (N); ++i) {
    cur -= vr[i];
    for (auto o : vl[i]) {
      cur++;
      Q.push(o);
    }
    while (cur > k) {
      cur--;
      ans.push_back(Q.top().second);
      vr[Q.top().first + 1]--;
      Q.pop();
    }
  }
  printf("%d\n", (int)(ans).size());
  for (int i = (0); i < ((int)(ans).size()); ++i)
    printf("%d%c", ans[i], " \n"[i == (int)(ans).size() - 1]);
  return 0;
}
int main() {
  while (~scanf("%d%d", &n, &k)) {
    Init();
    Solve();
  }
  return 0;
}