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>
const double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
using namespace std;
const int MOD = 1e9 + 7;
double gcd(double a, double b) { return a < 0.01 ? b : gcd(fmod(b, a), a); }
template <typename T>
T mymax(T a, T b) {
  return (a > b) ? a : b;
}
template <typename T>
T mymax(T a, T b, T c) {
  return mymax(a, mymax(b, c));
}
template <typename T>
T mymin(T a, T b) {
  return (a < b) ? a : b;
}
template <typename T>
T mymin(T a, T b, T c) {
  return mymin(a, mymin(b, c));
}
template <typename T>
T power(T x, T y) {
  T res = 1;
  x = x % MOD;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return res;
}
template <typename T>
void swap(T *x, T *y) {
  T temp;
  temp = *y;
  *y = *x;
  *x = temp;
}
template <typename T>
T mod(T a) {
  if (a > 0)
    return a;
  else
    return -a;
}
template <typename T>
T logg(T a) {
  T x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
typedef struct trie {
  int a, b, c;
  trie(int p1, int p2, int p3) { a = p1, b = p2, c = p3; }
} tri;
bool operator<(tri p, tri q) {
  if (p.a < q.a) {
    return 1;
  } else if (p.a > q.a)
    return 0;
  else {
    return (p.b < q.b);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k1;
  cin >> n >> k1;
  multiset<pair<int, int>> k;
  vector<int> l(n), r(n);
  vector<tri> p;
  for (int i = 1; i <= n; i++) {
    cin >> l[i - 1] >> r[i - 1];
    p.push_back({l[i - 1], -1, i - 1});
    p.push_back({r[i - 1], 1, i - 1});
  }
  int c = 0;
  sort(p.begin(), p.end());
  vector<int> ans;
  for (tri p1 : p) {
    int a1 = p1.a, b1 = p1.b, c1 = p1.c;
    if (b1 == -1) {
      k.insert({r[c1], c1});
      c++;
      if (c > k1) {
        pair<int, int> pk = *k.rbegin();
        k.erase(k.find(pk));
        ans.push_back(pk.second);
        c--;
      }
    } else {
      if (k.count({r[c1], c1})) {
        c--;
        k.erase(k.find({r[c1], c1}));
      }
    }
  }
  cout << ans.size() << '\n';
  for (int pk : ans) {
    cout << pk + 1 << ' ';
  }
  cout << '\n';
  return 0;
}