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 = 1e9 + 7;
struct DS {
  int bit[200005], n;
  int lowbit(int x) { return x & -x; }
  void add(int i, int d) {
    while (i <= n) {
      bit[i] += d;
      i += lowbit(i);
    }
  }
  void add(int l, int r, int d) {
    add(l, d);
    add(r + 1, -d);
  }
  int qry(int i) {
    int res = 0;
    while (i) {
      res += bit[i];
      i -= lowbit(i);
    }
    return res;
  }
} p;
pair<pair<int, int>, int> a[200005];
int32_t main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  p.n = 200004;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i + 1;
  }
  sort(a, a + n);
  set<pair<pair<int, int>, int>> st;
  for (int i = 0; i < n; i++) p.add(a[i].first.first, a[i].first.second, 1);
  vector<int> ans;
  for (int i = 1, j = 0; i < 200005; i++) {
    while (j < n && a[j].first.first == i) {
      st.insert({{a[j].first.second, a[j].first.first}, a[j].second});
      j++;
    }
    if (p.qry(i) > k) {
      int d = p.qry(i) - k;
      while (d--) {
        auto it = st.rbegin();
        int l = it->first.second, r = it->first.first;
        ans.push_back(it->second);
        p.add(l, r, -1);
        st.erase(*st.rbegin());
      }
    }
  }
  cout << ans.size() << endl;
  for (int e : ans) cout << e << " ";
}