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 maxn = 200005;
int n, k;
int s[maxn], seg[maxn], f[maxn];
vector<pair<int, int> > P[maxn];
void Solve(int Test) {
  cin >> n >> k;
  int mx = 0;
  for (int i = 1; i <= n; ++i) {
    int l, r;
    cin >> l >> r;
    mx = max(mx, r);
    P[l].push_back({r, i});
  }
  vector<int> ans;
  set<pair<int, int> > s;
  for (int i = 1; i <= mx; ++i) {
    while (s.size()) {
      auto it = s.begin();
      pair<int, int> cur = *it;
      if (cur.first < i)
        s.erase(it);
      else
        break;
    }
    for (auto x : P[i]) {
      s.insert(x);
    }
    while ((int)s.size() > k) {
      auto it = s.rbegin();
      pair<int, int> cur = *it;
      s.erase(s.find({cur.first, cur.second}));
      ans.push_back(cur.second);
    }
  }
  cout << ans.size() << "\n";
  for (auto x : ans) cout << x << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  if (fopen("CF"
            ".inp",
            "r")) {
    freopen(
        "CF"
        ".inp",
        "r", stdin);
    freopen(
        "CF"
        ".out",
        "w", stdout);
  }
  int test = 1;
  for (int i = 1; i <= test; ++i) {
    Solve(i);
  }
}