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 prac_1_1 {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int count = 0;
   //     boolean[] b_arr = new boolean[m+1];
   //     Arrays.fill(b_arr,false);
        int[][] mat = new int[n][2];
        for(int i=0;i<n;i++) {
            mat[i][0] = sc.nextInt();
            mat[i][1] = sc.nextInt();
            /*
            for(int j=mat[i][0];j<=mat[i][1];j++){
                if(!b_arr[j]) {
                    count--;
                    b_arr[j] = true;
                }
            }


        System.out.println(count);
        for (int k=1;k<=m;k++){
            if(!b_arr[k]){
                System.out.print(k+" ");
            }
        }
        */
        }
        Set<Integer> set = new HashSet<Integer>();
        boolean b = false;
        for(int var1=1;var1<=m;var1++){
            b=false;
            for (int var2=0;var2<n;var2++){
                if(var1>=mat[var2][0] && var1<=mat[var2][1]){
                    b=true;
                    continue;
                }
            }
            if(!b) {
                count++;
                set.add(var1);
            }
        }

        System.out.println(count);
        for(int vari:set){
            System.out.print(vari+" ");
        }
    }
}