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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) throws IOException {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader sc = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(1, sc, out);
        out.close();
    }

    static class Task {
    	public boolean[] take;
    	public Node[] buf;
    	
    	
    	public int find(int l,int n) {
    		int r=-1;
    		int index=-1;
    		
    		for(int i=1;i<=n;i++) {
    			if(take[i])
    				continue;
    			if(buf[i].l<=l) {
    				if(r<buf[i].r) {
    					r=buf[i].r;
    					index=i;
    				}
    			}
    		}
    		take[index]=true;
    		return index;
    	}
    	
        public void solve(int testNumber, InputReader sc, PrintWriter out) throws IOException {
        	int n=sc.nextInt();
        	int k=sc.nextInt();
        	buf=new Node[n+1];
        	take=new boolean[n+1];
        	ArrayList<Integer> res=new ArrayList<Integer>();
        	BIT bit=new BIT(200);
        	for(int i=1;i<=n;i++) {
        		int l=sc.nextInt();
        		int r=sc.nextInt();
        		buf[i]=new Node(l,r);
        		bit.add(l, r, 1);
        	}
        	for(int i=1;i<=200;i++) {
        		while(bit.ask(i)>k) {
        			int index=find(i,n);
        			res.add(index);
        			bit.add(buf[index].l, buf[index].r,-1);
        		}
        	}
        	out.println(res.size());
        	for(int v:res)
        		out.print(v+" ");
        	out.println();
        }

    }
    
    static class Node{
    	int l;
    	int r;
    	
    	public Node(int l,int r) {
    		this.l=l;
    		this.r=r;
    	}
    }
    
    static class BIT{
		public long[] C;
		
		public BIT(int n) {
			C=new long[n+1];
		}
		
		public int lowBit(int x) {
			return x&(-x);
		}
		
		public void add(int x,long y) {
			for(int i=x;i<C.length;i+=lowBit(i)) {
				C[i]+=y;
			}
		}
		
		public long ask(int x) {
			long res=0;
			
			for(int i=x;i>0;i-=lowBit(i))
				res+=C[i];
			
			return res;
		}
		
		//C[i]表示i这个数有C[i]个，相当于权值BIT，查询第k大。
	    public int kth_element(int k) {
	        int p = 0;
	        for (int lim = 1 << 20; lim>0; lim /= 2)
	            if (p + lim < C.length && C[p + lim] < k) {
	                p += lim;
	                k -= C[p];
	            }
	        return p + 1;
	    }
	    
	    public void add(int l,int r,long val) {
	    	this.add(l, val);
	    	this.add(r+1, -val);
	    }
	    
	    public long ask(int l,int r) {
	    	return this.ask(r)-this.ask(l-1);
	    }
	}

    static class InputReader{
        StreamTokenizer tokenizer;
        public InputReader(InputStream stream){
            tokenizer=new StreamTokenizer(new BufferedReader(new InputStreamReader(stream)));
            tokenizer.ordinaryChars(33,126);
            tokenizer.wordChars(33,126);
        }
        public String next() throws IOException {
            tokenizer.nextToken();
            return tokenizer.sval;
        }
        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }
        public boolean hasNext() throws IOException {
            int res=tokenizer.nextToken();
            tokenizer.pushBack();
            return res!=tokenizer.TT_EOF;
        }
        
        public double nextDouble() throws NumberFormatException, IOException {
        	return Double.parseDouble(next());
        }
        
        public BigInteger nextBigInteger() throws IOException {
        	return new BigInteger(next());
        }
    }
}