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;
using namespace std::chrono;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int mod = 1e9 + 7;
long long binpow(long long num, long long pow);
int extendedEuclid(int a, int b, int &x, int &y);
inline int mul(int a, int b) { return (a * 1ll * b) % mod; }
inline int add(int a, int b) { return (a + 0ll + b) % mod; }
inline int sub(int a, int b) { return ((a + 0ll - b) % mod + mod) % mod; }
inline void intarr(vector<int> &a) {
  for (auto &i : a) cin >> i;
}
inline void larr(vector<long long> &a) {
  for (auto &i : a) cin >> i;
}
const int N = 200200;
inline void solve() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> a(n);
  vector<int> cnt(N);
  vector<vector<int>> ev(N);
  for (int i = (int)0; i < (int)n; i++) {
    cin >> a[i].first >> a[i].second;
    cnt[a[i].first]++;
    cnt[a[i].second + 1]--;
    ev[a[i].first].emplace_back(i + 1);
    ev[a[i].second + 1].emplace_back(-i - 1);
  }
  for (int i = (int)1; i < (int)N; i++) cnt[i] += cnt[i - 1];
  vector<int> ans(n), sub(N);
  set<pair<int, int>> segs;
  int curr = 0;
  for (int i = (int)0; i < (int)N; i++) {
    curr += sub[i];
    for (auto it : ev[i]) {
      if (it > 0)
        segs.insert(make_pair(a[it - 1].second, it - 1));
      else {
        auto iter = segs.find(make_pair(a[-it - 1].second, -i - 1));
        if (iter != segs.end()) segs.erase(iter);
      }
    }
    while (cnt[i] - curr > k) {
      int pos = segs.rbegin()->second;
      segs.erase(prev(segs.end()));
      curr++;
      --sub[a[pos].second + 1];
      ans[pos] = 1;
    }
  }
  cout << accumulate(ans.begin(), ans.end(), 0) << endl;
  for (int i = (int)0; i < (int)n; i++)
    if (ans[i]) cout << i + 1 << " ";
  cout << endl;
}
int main() {
  auto start = high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop - start);
  cerr << duration.count() << " microseconds\n";
  return 0;
}
long long binpow(long long num, long long pow) {
  if (pow == 0) return 1;
  long long val = binpow(num, pow / 2);
  val *= val;
  if (pow & 1) val *= num;
  return val;
}
int extendedEuclid(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  int x1, y1;
  int d = extendedEuclid(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}