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.io.PrintWriter;
import java.util.*;

public class Main {

    public static void main(String[] args) {

        PrintWriter cout = new PrintWriter(System.out, true);
        Scanner cin = new Scanner(System.in);

        int n = cin.nextInt(), m = cin.nextInt(), start, end, p;
        List<Integer> points = new LinkedList<>();

        int[] tab = new int[m+1];

        for (int i = 0; i < n; i++) {
            start = cin.nextInt();
            end = cin.nextInt();

            for (int j = start; j <= end; j++) {
                tab[j] = 1;
            }
        }

        for (int i = 1; i < m+1; i++) {
            p = tab[i];
            if (p==0) {
                points.add(i);
            }
        }

        StringBuilder res = new StringBuilder(points.size() + "\n");

        for (Integer pt : points) {
            res.append(pt).append(" ");
        }

        cout.println(res.toString().trim());
    }
}