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 JAVA):

Solution:
import java.util.*;
import java.io.*;
public class D {

	static int N = 200000;
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		ArrayList<Pair>[] op = new ArrayList[N+1];
		for(int i = 0; i <= N; i++) {
			op[i] = new ArrayList<>();
		}
		for(int i = 0; i < n; i++) {
			st = new StringTokenizer(br.readLine());
			int a = Integer.parseInt(st.nextToken());
			int b = Integer.parseInt(st.nextToken());
			Pair p = new Pair(i+1, a, b);
			op[a].add(p);
		}
		TreeSet<Pair> currps = new TreeSet<>();
		ArrayList<Integer> res = new ArrayList<>();
		for(int i = 0; i <= N; i++) {
			while(!currps.isEmpty()) {
				Pair p = currps.last();
				if(p.t < i) currps.pollLast();
				else break;
			}
			currps.addAll(op[i]);
			while(currps.size() > k) {
				Pair p = currps.pollFirst();
				res.add(p.ind);
			}
		}
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		StringBuilder sb = new StringBuilder();
		for(int r: res) {
			sb.append(r+" ");
		}
		bw.write(res.size()+"\n");
		bw.write(sb.toString().trim()+"\n");
		bw.flush();
	}
	static class Pair implements Comparable<Pair>{
		int ind, s, t;
		public Pair(int ind, int s, int t) {
			this.ind = ind; this.s = s; this.t = t;
		}
		@Override
		public int compareTo(Pair p) {
			if(t == p.t && s == p.s) return ind - p.ind;
			if(t == p.t) return s - p.s;
			else return p.t - t;
		}
		public String toString() {
			return s+"-"+t;
		}
	}
}