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 long long int N = 5e5 + 5;
const long long int mod = 1e9 + 7;
const int di[4] = {-1, 0, 1, 0};
const int dj[4] = {0, -1, 0, 1};
const double pi = 3.14159265358979323846;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool vis[N];
vector<long long int> adj[N];
void dfs(long long int x) {
  vis[x] = true;
  for (int i = 0; i < adj[x].size(); i++) {
    if (!vis[adj[x][i]]) {
      dfs(adj[x][i]);
    }
  }
}
long long int dis[N], ct[N];
void bfs(long long int x) {
  queue<long long int> q;
  q.push(x);
  dis[x] = 1;
  ct[x] = 1;
  while (!q.empty()) {
    long long int t = q.front();
    q.pop();
    for (auto i : adj[t]) {
      if (!dis[i]) {
        dis[i] = dis[t] + 1;
        q.push(i);
        ct[i] = 1;
      } else if (dis[i] == dis[t] + 1)
        ct[i]++;
    }
  }
}
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  return a.second < b.second;
}
long long int po(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return po((x * x) % 998244353, n / 2);
  else
    return (x * po((x * x) % 998244353, (n - 1) / 2)) % 998244353;
}
long long int fact[N];
long long int ncr(long long int n, long long int r) {
  long long int d = fact[n];
  long long int e = (fact[n - r] * fact[r]) % 998244353;
  d *= po(e, 998244353 - 2);
  return d % 998244353;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  vector<pair<int, int> > v[200005];
  for (int i = 0; i < n; ++i) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back({y, i + 1});
  }
  set<pair<int, int> > s;
  vector<long long int> ans;
  for (int i = 1; i < 200005; ++i) {
    while (!s.empty() && s.begin()->first < i) s.erase(s.begin());
    for (int j = 0; j < v[i].size(); ++j) s.insert(v[i][j]);
    while (s.size() > k) {
      ans.push_back((--s.end())->second);
      s.erase(--s.end());
    }
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); ++i) cout << ans[i] << ' ';
  return 0;
}