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.*;
import java.lang.*;
import java.io.*;
public class TestClass {
   public static void main(String args[] ) throws Exception {
      Scanner s = new Scanner(System.in);
      int t=s.nextInt();
      int m=s.nextInt();
      int[] a = new int[m];
      int[] res = new int[m];
      int[][] l = new int[t][2];
      int p=1, j=0;
      int count=0;
      while((p<=m) && (j<m)){
          a[j++]= p;
          p++;
      }
      for(int i=0; i<t; i++){
          l[i][0] = s.nextInt();
          l[i][1] = s.nextInt();
      }
      int ind=0, flag=0;
      for(int i=0; i<m; i++){
          for(int k=0; k<t; k++){
              if(l[k][0]<=a[i] && l[k][1]>=a[i]){
                  flag=1;
                //   System.out.print(a[i] + " ");
                  break;
              }
              else{
                    flag=0;
                    // System.out.print(a[i] + " ");
              }
          }
          if(flag==0)
            res[ind] = a[i];
            if(res[ind++]!=0)
                count++;
      }
      System.out.println(count);
      for(int i=0; i<m; i++){
          if(res[i] != 0)
            System.out.print(res[i] + " ");
      }
    }
}

