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.*;
import java.io.*;
public class codeforces {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int size = scan.nextInt();

        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = scan.nextInt();
            b[i] = scan.nextInt();
        }
        boolean[] c = new boolean[size+1];
        for (int i = 1; i <= size; i ++) {
            for (int p = 0; p < n; p++) {
                if (i <= b[p] && i>= a[p])
                    c[i] = true;
            }
        }
        int count = 0;
        ArrayList hi = new ArrayList<Integer>();
        for (int i = 1; i <= size; i++) {
            if (!c[i]) {
                hi.add(i);
                count++;
            }
        }

        if (count == 0)
            System.out.println(0);
        else {
            System.out.println(count);
            for (int i = 0; i < hi.size(); i++)
                System.out.print(hi.get(i) + " ");
        }

    }
}