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;

public class Main {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		/** 1 **/
		int n = scan.nextInt();	// 세그먼트 집합 수
		int m = scan.nextInt();	// 정수의 끝 점
		boolean[] check = new boolean[m+1];
		for(int i=0; i<n; i++) {
			int l = scan.nextInt();	// 왼쪽 점
			int r = scan.nextInt();	// 오른쪽 점
			// 해당 범위(l ~ r)는 세그먼트에 속하므로 체크
			for(int j=l; j<=r; j++) 
				check[j] = true;
			
		}
		
		int dot = 0;	// 어떤 세그먼트에 속하지 않는 점의 수
		for(int i=1; i<check.length; i++) {
			if(!check[i])	// l, r의 해당점에 의해 체크가 안된 경우
				dot ++;
		}
		
		System.out.println(dot);
		if(dot == 0)
			return;
		
		for(int i=1; i<check.length; i++)
			if(!check[i])
				System.out.print(i + " ");
		
		scan.close();
	}

}
