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 sol {
	public static void main(String args[]) {
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int k = input.nextInt();
		int arr[] = new int[202];
		ArrayList<pair> al = new ArrayList<>();
		int arr_pair[] = new int[n];
		int ans = 0;
		for(int i=0;i<n;i++) {
			int l = input.nextInt();
			int r = input.nextInt();
			arr[l]++;
			arr[r+1]--;
			pair temp = new pair(l,r);
			al.add(temp);
			arr_pair[i]++;
		}
		for(int i=1;i<202;i++) {
			arr[i] += arr[i-1];
		}
		for(int i=0;i<202;i++) {
			while(arr[i]>k) {
				int pair_index = -1;
				for(int j=0;j<n;j++) {
					pair temp = al.get(j);
					if(arr_pair[j]!=0 && temp.l<=i && temp.r>=i && (pair_index == -1 || temp.r> al.get(pair_index).r)) {
						pair_index = j;
					}
				}

				arr_pair[pair_index]--;
				ans++;
				for(int itr=al.get(pair_index).l;itr<=al.get(pair_index).r;itr++) {
					arr[itr]--;
				}
			}

		}
		System.out.println(ans);
		for(int i=0;i<n;i++) {
			if(arr_pair[i]==0) {
				System.out.print((i+1)+" ");
			}
		}
	}
}
class pair{
	int l;
	int r;
	pair(int l,int r){
		this.l = l;
		this.r = r;
	}
	public boolean equals(Object other) {
		 if(!(other instanceof pair)) {
		        return false;
		   }
		 pair that = (pair)other;
		 return this.l == that.l && this.r==that.r;
	}
	public int hashCode() {
	    int hashCode = 1;

	    hashCode = hashCode * 37 + this.l;
	    hashCode = hashCode * 37 + this.r;

	    return hashCode;
	}
	public String toString() {
		return this.l+" "+this.r;
	}
}