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 = 1e6 + 10, maxm = 2e6 + 10;
const long long INF = 0x3f3f3f3f, mod = 1e9 + 7;
int casn, m, k;
unsigned long long n;
class heap {
 public:
  priority_queue<pair<long long, long long> > x, y;
  void push(pair<long long, long long> a) { x.push(a); }
  void remove(pair<long long, long long> a) { y.push(a); }
  pair<long long, long long> top() {
    while (y.size() && x.top() == y.top()) x.pop(), y.pop();
    return x.top();
  }
  int size() { return x.size() - y.size(); }
  void pop() {
    while (y.size() && x.top() == y.top()) x.pop(), y.pop();
    x.pop();
  }
  pair<long long, long long> sectop() {
    pair<long long, long long> a = top();
    pop();
    pair<long long, long long> b = top();
    push(a);
    return b;
  }
} que;
vector<pair<long long, long long> > vis[maxn];
int f[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    int a, b;
    cin >> a >> b;
    vis[a].push_back(make_pair(b, i));
    vis[b + 1].push_back(make_pair(b, -i));
  }
  vector<int> ans;
  int cnt = 0;
  for (int i = 1; i <= 2e5; ++i) {
    for (pair<long long, long long> j : vis[i]) {
      if (j.second > 0) {
        que.push(j);
      } else if (j.second < 0) {
        if (!f[-j.second]) que.remove(make_pair(j.first, -j.second));
      }
    }
    while (que.size() > k) {
      ans.push_back(que.top().second);
      f[que.top().second] = 1;
      que.pop();
    }
  }
  cout << ans.size() << endl;
  for (auto i : ans) cout << i << ' ';
  cout << endl;
}