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;
int n, m, k, h;
long long pref[2 * 100000 + 5];
unordered_map<int, int> map_val;
vector<int> value;
vector<int> doNeed;
int modulo = 1000000007;
int arr[1000000 + 5];
int t[1000000];
void build(int a[], int v, int tl, int tr) {
  if (tl == tr)
    t[v] = a[tl];
  else {
    int tm = (tl + tr) / 2;
    build(a, v * 2, tl, tm);
    build(a, v * 2 + 1, tm + 1, tr);
  }
}
void update(int v, int tl, int tr, int l, int r, int add) {
  if (l > r) return;
  if (l == tl && tr == r)
    t[v] += add;
  else {
    int tm = (tl + tr) / 2;
    update(v * 2, tl, tm, l, min(r, tm), add);
    update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, add);
  }
}
int get(int v, int tl, int tr, int pos) {
  if (tl == tr) return t[v];
  int tm = (tl + tr) / 2;
  if (pos <= tm)
    return t[v] + get(v * 2, tl, tm, pos);
  else
    return t[v] + get(v * 2 + 1, tm + 1, tr, pos);
}
vector<pair<int, bool>> all_nodes;
int main() {
  cin >> n >> k;
  vector<pair<pair<int, int>, int>> all_vert(n);
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    all_vert[i] = make_pair(make_pair(a, b), i);
    update(1, 0, 210000, a, b, 1);
  }
  sort(all_vert.begin(), all_vert.end());
  vector<int> res;
  priority_queue<pair<int, pair<int, int>>> q;
  int vertInd = 0;
  for (int i = 0; i < 2100000; i++) {
    int cur = get(1, 0, 210000, i);
    if (cur > k) {
      pair<int, pair<int, int>> cur_p;
      for (int j = vertInd; j < n; j++) {
        if (all_vert[j].first.first <= i) {
          q.push(make_pair(
              all_vert[j].first.second,
              make_pair(all_vert[j].first.first, all_vert[j].second)));
          if (j == n - 1) {
            vertInd = n;
          }
        } else {
          vertInd = j;
          break;
        }
      }
      cur_p = q.top();
      q.pop();
      update(1, 0, 210000, cur_p.second.first, cur_p.first, -1);
      i--;
      res.push_back(cur_p.second.second + 1);
    }
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); i++) {
    cout << res[i] << " ";
  }
}