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 ≤ 200) — 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 ≤ 200) — 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
1 4 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 2 4 


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:
// No sorceries shall previal. //         
import java.util.Scanner;
import java.io.PrintWriter;
import java.util.Arrays;

public class InVoker { 
	
	static class Plate {
		int left;
		int right;
		int size;
		Plate(int l, int r){
			left=l;
			right=r;
			size=r-l+1;
		}
	}
	
	static void removePlate(Plate p, int fre[]) {
		for(int i=p.left;i<=p.right;i++) {
			fre[i]--;
		}
	}
	static void AddPlate(Plate p, int fre[]) {
		for(int i=p.left;i<=p.right;i++) {
			fre[i]++;
		}
	}
	static int findMaxLenIndex(Plate p[], boolean deleted[], int n,int cur) {
		int maxLen=0;
		int index=-1;
		for(int i=0;i<n;i++) {
			if(deleted[i])
				continue;
			if(maxLen < p[i].right-cur+1 && p[i].left<=cur+1) {
				index=i;
				maxLen=p[i].right-cur+1;
			}
		}
		return index;
	}
	
    public static void main(String args[]) {
    		
    	Scanner inp=new Scanner(System.in);
    	PrintWriter out= new PrintWriter(System.out);
    	
    	int n=inp.nextInt();
    	int k=inp.nextInt();
    	Plate p[]=new Plate[n];
    	int freq[]=new int[201];
    	boolean deleted[]=new boolean[n];
    	for(int i=0;i<n;i++) {
    			p[i]=new Plate(inp.nextInt(),inp.nextInt());
    			AddPlate(p[i],freq);
    	}
    	int gg=0;
    	for(int i=0;i<200;i++) {
    		if(freq[i+1]>k) {
    			//System.out.println(i);
    			int index=findMaxLenIndex(p,deleted,n,i);
    			removePlate(p[index],freq);
    			deleted[index]=true;
    			out.print(index+1+" ");
    			gg++;
    			i--;
    		}
    	}
    	
    	System.out.println(gg);
    	inp.close();
    	out.close();
    }    		
}