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 A{
     public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt(), t = s.nextInt();
        
        int sum = 0;
        int ara[] = new int[t];
        int ara1[] = new int[n];
        int ara2[] = new int[n];

        for(int i = 0; i < n; i++) {
            ara1[i] = s.nextInt();
            ara2[i] = s.nextInt();
        }

        for(int j = 0; j < n; j++) {
            int swap;
            for(int k = 0; k < n-1; k++) {

                if(ara1[k]>ara1[k+1]) {
                    swap = ara1[k+1];
                    ara1[k+1] = ara1[k];
                    ara1[k] = swap;
                    swap = ara2[k+1];
                    ara2[k+1] = ara2[k];
                    ara2[k] = swap;
                }
            }
        }

        int i = 0;
        int l = 1;

        for(int m = 0; m < n; m++) {

            for(int x = l; x < t; ) {
                if(x<=ara2[m]&&!(ara1[m]<=x&&ara2[m]>=x)) {
                    sum++;
                    ara[i] = x;
                    i++;

                }
                if(x>ara2[m]) {

                    break;

                }
                x++;
                l = x;

            }
        }
        Arrays.sort(ara2);
        System.out.println((t-ara2[n-1])+sum);

        for(int in : ara) {
            if(in==0)
                break;
            System.out.print(in+" ");
        }
        for(int v = ara2[n-1]+1; v <= t; v++){
            System.out.print(v+" ");
        }

    }
}