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.*;
public class cf1
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int n;int l;int i=0;int x,y,z;int c=0;int c1=0;int c2=0;int gg=0;String ans="";
        
        n=sc.nextInt();
        int ar[]=new int[2*n];
        
        l=sc.nextInt();
        
        for(i=0;i<2*n;i++)
        {
            ar[i]=sc.nextInt();
            if(i%2==0)
            System.out.println();
        }
        for(x=1;x<=l;x++)
        {c2=0;
            for(y=0;y<n;y++)
            {c=0;c1=0;
                for(z=ar[2*y];z<=ar[2*y+1];z++)
                {   c1++;
                    if(z!=x)
                    c++;
                }
                 if(c==c1)
                 c2++;
                }
             if(c2==n)
{
             ans=ans+" "+String.valueOf(x);
             gg++;
}
}
System.out.println(gg);
System.out.println(ans);
        
    }
}