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.*;
import java.io.*;
public class B{
	public static void main(String[] args)
	{
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = fs.nextInt(); int k = fs.nextInt();
		int[] power = new int[205];
		int[][] store = new int[n][4];
		for(int i=0;i<n;i++)
		{
			int l = fs.nextInt(); int r = fs.nextInt();
			for(int j=l;j<=r;j++)
			{
				power[j]++;
			}
			store[i][0] = (i+1);
			store[i][1] = l;
			store[i][2] = r;
		}
		int ans = 0;
		List<Integer> ret = new ArrayList();
		for(int i=1;i<=200;i++)
		{
			if(power[i]>k)
			{
				List<int[]> temp = new ArrayList();
				for(int j=0;j<n;j++)
				{
					if(store[j][3] == 1)continue;
					int left = store[j][1]; int right = store[j][2];
					if(i>=left && i<=right)
					{
						int[] add = new int[] {store[j][0],left,right};
						temp.add(add);
					}
				}
				Collections.sort(temp,(a,b)->{
					int ra = a[2];
					int rb = b[2];
					if(ra>rb)return -1;
					else if(rb>ra)return 1;
					else return 0;
 				});
				while(power[i]>k)
				{
					int idx = temp.get(0)[0]-1;
					store[idx][3] = 1;
					ans++;
					ret.add(idx+1);
					int left = store[idx][1]; int right = store[idx][2];
					for(int place=left;place<=right;place++)power[place]--;
					temp.remove(0);
				}
			}
		}
		out.println(ans);
		for(int i:ret)out.print(i+" ");
		out.println();
		
		out.close();
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	public static int[] sort(int[] arr)
	{
		List<Integer> temp = new ArrayList();
		for(int i:arr)temp.add(i);
		Collections.sort(temp,Collections.reverseOrder());
		int start = 0;
		for(int i:temp)arr[start++]=i;
		return arr;
	}
}