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>
#pragma GCC diagnostic ignored "-Wunused-result"
#pragma GCC diagnostic ignored "-Wunused-function"
using namespace std;
namespace {
namespace unolight {
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%" PRId64, &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%" PRId64, x); }
void _W(const double &x) { printf("%.16f\n", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
template <class T>
inline bool chmax(T &a, const T &b) {
  return b > a ? a = b, true : false;
}
template <class T>
inline bool chmin(T &a, const T &b) {
  return b < a ? a = b, true : false;
}
template <class T>
using MaxHeap = priority_queue<T>;
template <class T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <class T, class F = less<T>>
void sort_uniq(vector<T> &v, F f = F()) {
  sort(begin(v), end(v), f);
  v.resize(unique(begin(v), end(v)) - begin(v));
}
const int N = 200010;
int l[N], r[N];
void main() {
  int n, k;
  R(n, k);
  for (int i = 0; i < int(n); i++) {
    R(l[i], r[i]);
  }
  vector<int> id(n);
  iota(begin(id), end(id), 0);
  sort(begin(id), end(id), [&](int i, int j) {
    if (l[i] == l[j]) return r[i] > r[j];
    return l[i] < l[j];
  });
  for (int i = 0; i < int(n); i++)
    ;
  set<pair<int, int>> s;
  vector<int> ans;
  int j = 0;
  for (int i = 1; i < N; i++) {
    while (((int)(s).size()) && s.begin()->first < i) s.erase(s.begin());
    while (j < n && l[id[j]] <= i) {
      s.emplace(r[id[j]], id[j]);
      j++;
    }
    while (((int)(s).size()) > k) {
      auto it = --s.end();
      ans.push_back(it->second + 1);
      s.erase(it);
    }
  }
  W(ans.size());
  W(ans);
}
}  // namespace unolight
}  // namespace
int main() {
  unolight::main();
  return 0;
}