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.
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/**
 *
 * @author Sethu
 */
import java.util.Scanner;
public class Con500 {
    public static void main(String args[])
    {
        Scanner s=new Scanner(System.in);
        int n=s.nextInt(),m=s.nextInt();
        int a[]=new int[2*n];
        int b[]=new int[100];
        int bi=0;
        for(int i=0;i<2*n;i++)
        a[i]=s.nextInt();
        
        for(int i=1;i<=m;i++)
        {boolean t=false;
            for(int j=0;j<2*n;j+=2)
                if(i>=a[j]&&i<=a[j+1])
                    t=true;
            if(!t)
               b[bi++]=i;
         }
    if(bi>0)
    {  System.out.println(bi);
        for(int i=0;i<bi;i++)
            System.out.print(b[i]+" ");
    }
    else
        System.out.print("0");
    }
    
}
