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 Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=1000000007;
	long gcd(long a,long b) {
		return b==0?a:gcd(b,a%b);
	}
	void work() {
		int n=in.nextInt();
		int k=in.nextInt();
		ArrayList<Integer> ret=new ArrayList<>();
		int[][] A=new int[n][];
		for(int i=0;i<n;i++) {
			int s=in.nextInt();
			int e=in.nextInt();
			A[i]=new int[] {s,e,i+1};
		}
		Arrays.sort(A,new Comparator<int[]>() {
			public int compare(int[] arr1,int[] arr2) {
				if(arr1[0]==arr2[0]) {
					return arr2[1]-arr1[1];
				}
				return arr1[0]-arr2[0];
			}
		});
		TreeMap<Integer,Stack<Integer>> map=new TreeMap<>();
		for(int i=0,cnt=0;i<n;) {
			if(map.size()==0||A[i][0]<=map.firstKey()) {
				if(cnt<k) {
					if(map.get(A[i][1])==null)map.put(A[i][1],new Stack<Integer>());
					map.get(A[i][1]).add(A[i][2]);
					cnt++;
				}else {
					int key=map.lastKey();
					if(key<A[i][1]) {
						ret.add(A[i][2]);
					}else {
						ret.add(map.get(key).pop());
						if(map.get(key).size()==0) map.remove(key);
						if(map.get(A[i][1])==null)map.put(A[i][1],new Stack<Integer>());
						map.get(A[i][1]).add(A[i][2]);
					}
				}
				i++;
			}else {
				int key=map.firstKey();
				map.get(key).pop();
				if(map.get(key).size()==0)map.remove(key);
				cnt--;
			}
		}
		out.println(ret.size());
		for(int r:ret) {
			out.print(r+" ");
		}
	}
}



class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}

	public String next() 
	{
		if(st==null || !st.hasMoreElements())
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	public int nextInt() 
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}