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 = 2e5 + 10;
pair<pair<int, int>, int> node[maxn];
struct cmp {
  bool operator()(const pair<int, int> p1, const pair<int, int> p2) {
    return p1.first < p2.first;
  }
};
priority_queue<int, vector<int>, greater<int> > les;
priority_queue<pair<int, int>, vector<pair<int, int> >, cmp> gret;
vector<int> vi;
int num[maxn];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    node[i] = make_pair(make_pair(x, y), i);
  }
  sort(node, node + n);
  int now = 0;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    now += 1;
    les.push(node[i].first.second);
    gret.push(pair<int, int>(node[i].first.second, node[i].second));
    while (les.top() < node[i].first.first) {
      if (!num[les.top()])
        now--;
      else
        num[les.top()]--;
      les.pop();
    }
    if (now > k) {
      ans++;
      num[gret.top().first]++;
      vi.push_back(gret.top().second);
      gret.pop();
      now--;
    }
  }
  cout << ans << endl;
  sort(vi.begin(), vi.end());
  for (int i = 0; i < ans; i++) {
    cout << vi[i] + 1 << ' ';
  }
}