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.util.*;
         
        public class axis{
            public static void main(String[] args){
                Scanner S= new Scanner(System.in);
                int n= S.nextInt();
                int m =S.nextInt();
                
                int a[][] = new int[n][2];
                
                for(int i=0;i<n;i++)
                {
                    for(int j=0;j<2;j++)
                    {
                        a[i][j]=S.nextInt();
                    }
                }
                
                int b[] = new int[m+1];
                for(int i=0;i<n;i++)
                {
                    for(int j=a[i][0];j<=a[i][1];j++)
                    {
                        b[j]++;
                    }
                }
                int c=0;
                for(int i=1;i<=m;i++)
                {
                    if(b[i]==0)
                    c++;
                }
                System.out.println(c);
                
                for(int i=1;i<=m;i++)
                {
                    if(b[i]==0)
                    System.out.print(i+" ");
                
                    
                }
                
                
            }
        }