You are given a set of n segments on the axis Ox, each segment has integer endpoints between 1 and m inclusive. Segments may intersect, overlap or even coincide with each other. Each segment is characterized by two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — coordinates of the left and of the right endpoints. 

Consider all integer points between 1 and m inclusive. Your task is to print all such points that don't belong to any segment. The point x belongs to the segment [l; r] if and only if l ≤ x ≤ r.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100) — the number of segments and the upper bound for coordinates.

The next n lines contain two integers each l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — the endpoints of the i-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that l_i=r_i, i.e. a segment can degenerate to a point.

Output

In the first line print one integer k — the number of points that don't belong to any segment.

In the second line print exactly k integers in any order — the points that don't belong to any segment. All points you print should be distinct.

If there are no such points at all, print a single integer 0 in the first line and either leave the second line empty or do not print it at all.

Examples

Input

3 5
2 2
1 2
5 5


Output

2
3 4 


Input

1 7
1 7


Output

0

Note

In the first example the point 1 belongs to the second segment, the point 2 belongs to the first and the second segments and the point 5 belongs to the third segment. The points 3 and 4 do not belong to any segment.

In the second example all the points from 1 to 7 belong to the first segment.
import java.util.*;
import java.io.*; 
import java.text.*;

public class Main{
    long mod = (int)1e9+7, IINF = (long)1e17;
    final int MAX = (int)1e3+1, INF = (int)1e9, root = 3;
    DecimalFormat df = new DecimalFormat("0.00000000");
    double PI = 3.141592653589793238462643383279502884197169399375105820974944;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }

    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        for(int i = 1, T= (multipleTC)?ni():1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    
    void solve(int TC) throws Exception{
        int n = ni(), m = ni();
        int[] sum = new int[m+2];
        for(int i = 0; i< n; i++){
            sum[ni()]++;
            sum[ni()+1]--;
        }
        for(int i = 1; i<= m; i++)sum[i]+=sum[i-1];
        int cnt = 0;
        StringBuilder ans = new StringBuilder("");
        for(int i = 1; i<= m; i++){
            if(sum[i]==0){
                cnt++;
                ans.append(i+" ");
            }
        }
        pn(cnt+"\n"+ans.toString());
    }
    
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{    
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }   
            return str;
        }
    }
}     