Problem: 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.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class C {
    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[] nextArray(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());
        }
    }
    static class FastWriter extends PrintWriter {
        FastWriter() {
            super(System.out);
        }

        void println(int[] array) {
            for (int i = 0; i < array.length; i++) {
                print(array[i] + " ");
            }
            println();
        }

        void println(long[] array) {
            long ans[]=new long[10];
            for (int i = 0; i < array.length; i++) {
                print(array[i] + " ");
            }
            println();
        }
    }
    public static void main(String[] args) {
        // Yo bitch !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
        FastScanner in = new FastScanner();
        FastWriter out = new FastWriter();
        int n=in.nextInt();
        int m=in.nextInt();
        int[] ar=new int[m+1];
        for (int i = 0; i < n; i++) {
            int l=in.nextInt();
            int r=in.nextInt();
            for (int j = l; j <= r; j++) {
                ar[j]=1;
            }
        }
        int cnt=0;
        for (int i = 1; i < m + 1; i++) {
            if(ar[i]==0){
                cnt++;
            }
        }
        out.println(cnt);
        for (int i = 1; i < m + 1; i++) {
            if(ar[i]==0){
                out.print(i+" ");
            }
        }
        out.println();
        out.close();
    }
}