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 N = 200 * 1000 + 17;
const int MOD = 1000 * 1000 * 1000 + 7;
int n, k;
vector<pair<int, int> > rg[N];
bool used[N];
bool read() {
  if (!(cin >> n >> k)) return false;
  for (int i = 0; i < n; ++i) {
    int l, r;
    cin >> l >> r;
    rg[l].emplace_back(r, i);
  }
  return true;
}
void solve() {
  set<pair<int, int> > E;
  for (int i = 0; i < N; ++i) {
    while (!E.empty() && E.begin()->first < i) E.erase(E.begin());
    for (int j = 0; j < rg[i].size(); ++j) {
      E.insert(rg[i][j]);
      used[rg[i][j].second] = true;
    }
    while (E.size() > k) {
      auto it = --E.end();
      used[it->second] = false;
      E.erase(it);
    }
  }
  vector<int> ans;
  for (int i = 0; i < n; ++i)
    if (!used[i]) ans.emplace_back(i);
  cout << ans.size() << endl;
  for (auto i : ans) cout << i + 1 << ' ';
  cout << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  while (read()) solve();
  return 0;
}