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.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
  
    void solve() throws IOException {
        int n=ni(),k=ni();
        int[][]A=new int[n+1][2];
        int[]B=new int[n+1];
        for (int x=0;x<n;x++) {
            A[x][0]=ni();
            A[x][1]=x+1;
            B[x+1]=ni();
        }
        A[n][0]=Integer.MAX_VALUE;
        Arrays.sort(A,(a,b)->Integer.compare(a[0],b[0]));
        
        LinkedList<Integer>C = new LinkedList();
        LinkedList<Integer>D = new LinkedList();
        ArrayList<Integer>F=new ArrayList();
        
        int p=0;
        int q=0;
        while (p<n) {
            q=A[p][0];
            while (C.size()>0 && C.get(0)<q) { C.remove(0); D.remove(0); }
            while (A[p][0]==q) {
                int nv=A[p][1];
                int a=Collections.binarySearch(C,B[nv]);
                if (a<0) a=(a+1)*-1;
                if (a==C.size()) { C.add(B[nv]); D.add(nv); }
                else { C.add(a,B[nv]); D.add(a,nv); }
                p++;
            }
            while (C.size()>k) {
                F.add(D.get(D.size()-1));
                C.remove(C.size()-1);
                D.remove(D.size()-1);
            }
        }
        
        out.println(F.size());
        for (int x=0;x<F.size();x++) out.print(F.get(x)+" ");
        out.println();
        out.flush();
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}