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:
import java.util.*;

public class d2{
    static int N = 200000;
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int k = scan.nextInt();
        pair[] pairs = new pair[n];
        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++){
            int a = scan.nextInt();
            int b = scan.nextInt();
            // pairs[i] = new pair(a,b,i+1);
            op[a].add(new pair(b,i+1));
        }
        // Collections.sort(pairs,new Comparator<pair>(){
        //     public int compare(pair a, pair b)
        //     {
        //         if(a.l==b.l && a.r==b.r)
        //             return a.i-b.i;
        //         else if (a.l==b.l)
        //             return a.r-b.r;
        //         else
        //             return a.l-b.l;
        //     }
        // });
        TreeSet<pair> tset = new TreeSet<pair>();
        ArrayList<Integer> ans = new ArrayList<>();
        for(int i=0;i<N+1;i++)
        {
            
            for(pair x:op[i])
                tset.add(x);
            // System.out.println(tset.size()+" "+op[i].size());
            while(!tset.isEmpty() && tset.first().r<i)
            {
                tset.pollFirst();
            }

            while(tset.size()>k)
            {
                ans.add(tset.last().i);
                tset.pollLast();
            }
        }

        System.out.println(ans.size());
        for(int i:ans)
            System.out.print(i+" ");
        System.out.println();
        
    }
}
class pair implements Comparable<pair>{
    int r,i;
    public pair(int r, int i)
    {
        this.r = r;
        this.i = i;
    }
    // @override
    public int compareTo(pair p) {
			if(r == p.r) return i - p.i;
			else return r-p.r;
		}
}