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 codeforce
{
	public static void main(String[] args) 
	{

		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		int m=s.nextInt();
		boolean b[]=new boolean[m];
		Arrays.fill(b,false);
		int count=m;
		for(int i=0;i<n;i++)
		{
			int l=s.nextInt()-1;
			int r=s.nextInt()-1;
			while(l<=r)
			{
				if(b[l]&b[r])
					{l++;r--;}
				 else
				 { if(!b[l])
				 				 {
				 				 	count--;
				 				 	//System.out.println("i "+l+" true");
				 				 	b[l]=true;
				 				 	l++;
				 				 }
				 				 if(!b[r])
				 				 {
				 				 	if((l-1)!=(r))
				 				 		count--;
				 				 	//System.out.println("i "+l+" true");
				 				 	b[r]=true;
				 			 		r--;
				 				 }}
			}
		}
		if(count!=0)
		{
			System.out.println(count);
			for(int i=0;i<m;i++)
			{
				if(!b[i])
				{
					System.out.print(i+1+" ");
				}
			}
			System.out.println();
		}
		else
		{
			System.out.println(0);
		}
	}
}