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 PointsInSegment{
	public static void main(String []args){
		
		java.util.Scanner scan=new java.util.Scanner(System.in);
		int n=scan.nextInt();
		int m=scan.nextInt();
		int [][]arr=new int[n][2];
		for(int i=1;i<=arr.length;i++){
			for(int j=1;j<=2;j++){
				arr[i-1][j-1]=scan.nextInt();
			}
		}
		HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
		for(int i=1;i<=m;i++){
			map.put(i,1);
		}
		for(int i=1;i<=arr.length;i++){
			for(int j=arr[i-1][0];j<=arr[i-1][1];j++){
				if(map.containsKey(j)){
					map.remove(j,1);
				}
			}
		}
		ArrayList<Integer> list=new ArrayList<Integer>();
		Iterator hmIterator = map.entrySet().iterator(); 
  
        while (hmIterator.hasNext()) { 
            Map.Entry mapElement = (Map.Entry)hmIterator.next(); 
            list.add((int)mapElement.getKey());
        } 
		Collections.sort(list);
		System.out.println(list.size());
		for(int i=1;i<=list.size();i++){
			System.out.print(list.get(i-1)+" ");
		}
	}
}