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 Solution {
	public static void main (String[] args)throws IOException{
	    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String l[] = br.readLine().split(" ");
        int n = Integer.parseInt(l[0]);
        int k = Integer.parseInt(l[1]);
        int arr[][] = new int[n][2];
        int t = 200001;
        int val[] = new int[t];
        HashMap<Integer,LinkedList<Integer>> hm = new HashMap<>();
        for(int i=0;i<n;i++)
        {
            String g[] = br.readLine().split(" ");
            int a = Integer.parseInt(g[0]);
            int b = Integer.parseInt(g[1]);
            arr[i][0] = a;
            arr[i][1] = b;
            val[a-1]++;
            val[b]--;
            LinkedList<Integer> lis = hm.getOrDefault(a-1,new LinkedList<Integer>());
            lis.add(i);
            hm.put(a-1,lis);
            lis = hm.getOrDefault(b,new LinkedList<Integer>());
            lis.add(-i);
            hm.put(b,lis);
        }
        StringBuilder sb = new StringBuilder();
        int sum = 0;
        int len=0;
        TreeMap<Integer,TreeSet<Integer>> h = new TreeMap<>();
        for(int i=0;i<t;i++)
        {
            sum += val[i];
            LinkedList<Integer> f = ((hm.containsKey(i))?hm.get(i):null);
            if(f!=null){
            for(int x:f)
            {
                if(x<0)
                {
                    TreeSet<Integer> hs =  h.get(arr[-x][1]);
                    if(hs!=null && hs.contains(-x))hs.remove(-x);
                }
                else
                {
                    if(h.containsKey(arr[x][1]))
                    {
                        TreeSet<Integer> hs =  h.get(arr[x][1]);
                        hs.add(x);
                    }
                    else
                    {
                        TreeSet<Integer> hs = new TreeSet<>();
                        hs.add(x);
                        h.put(arr[x][1],hs);
                    }
                }
            }
            }
                while(sum-k>0)
                {
                    int rm = h.lastKey();
                    TreeSet<Integer> hs = h.get(rm);
                    int y = hs.pollLast();
                    sum--;
                    val[rm]++;
                    if(hs.size()==0)h.remove(rm);
                    len++;
                    sb.append(y+1+" ");
                }
	    }
	    System.out.println(len);
        System.out.println(sb);
	}
}