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 dx[8] = {-1, 0, 0, 1, 1, 1, -1, -1};
const int dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};
long long axis[200005] = {0};
long long axis1[200005] = {0};
vector<pair<int, int>> maxp[200005];
void solve() {
  int n, k;
  cin >> n >> k;
  int i;
  int l, r;
  for (i = 0; i < n; i++) {
    cin >> l >> r;
    axis[l] += 1;
    axis[r + 1] -= 1;
    maxp[l].push_back({r, i + 1});
  }
  long long sum = 0;
  for (i = 0; i < 200005; i++) {
    sum += axis[i];
    axis[i] = sum;
  }
  sum = 0;
  vector<int> answers;
  priority_queue<pair<int, int>> moves;
  for (i = 0; i < 200005; i++) {
    if (!maxp[i].empty()) {
      for (auto u : maxp[i]) {
        moves.push(u);
      }
    }
    sum += axis1[i];
    axis[i] += sum;
    while (axis[i] > k) {
      auto x = moves.top();
      moves.pop();
      axis1[x.first + 1] += 1;
      answers.push_back(x.second);
      axis1[i + 1] -= 1;
      axis[i]--;
    }
  }
  cout << answers.size() << endl;
  for (auto u : answers) {
    cout << u << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  for (int k = 1; k <= tc; k++) {
    solve();
  }
  return 0;
}