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.Scanner;
import java.util.ArrayList;

public class Segments {
	public static void main(String[] args) {
		
		Scanner in = new Scanner(System.in);		
		int n = in.nextInt(), m = in.nextInt(), i;
		ArrayList<Integer> a = new ArrayList<Integer>();
		
		for (i=1; i<m+1; i++) { a.add(i); }
		
		for (i=0; i<n; i++) {
			
			int l = in.nextInt(), r = in.nextInt(), j;

			for (j=l; j<r+1; j++) {
				if (a.contains(j)) { a.remove(a.indexOf(j)); }
			}
		}
		System.out.println(a.size());
		for (i=0; i<a.size(); i++) { System.out.print(a.get(i) + " "); }
		
		in.close();
	}
}
