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.*;
import java.util.*;

public class Main {
   private void solve() {
      int kol = nextInt();
      int gran = nextInt();
      boolean[] prin = new boolean[gran];
      int ans = 0;
      for(int i = 0; i < kol; i++) {
         int f = nextInt();
         int s = nextInt();
         for(int j = f; j <= s; j++) {
               prin[j - 1] = true;
         }
      }
      for(int i = 0; i < gran; i++) {
         if(prin[i] == false) {
            ans++;
         }
      }
      out.println(ans);
      for(int i = 0; i < gran; i++) {
         if(prin[i] == false)
            out.print(i + 1 + " ");
      }
   }

   public static void main(String[] args) {
      new Main().run();
   }

   BufferedReader br;
   StringTokenizer st;
   PrintWriter out;

   private void run() {
      try {
         br = new BufferedReader(new InputStreamReader(System.in));
         out = new PrintWriter(System.out);
         // br = new BufferedReader(new FileReader("birthday.in"));
         // out = new PrintWriter(new FileWriter("birthday.out"));
         solve();
         br.close();
         out.close();
      } catch (IOException e) {
         e.printStackTrace();
      }
   }

   private int nextInt() {
      return Integer.parseInt(next());
   }

   private String next() {
      while (st == null || !st.hasMoreTokens()) {
         try {
            st = new StringTokenizer(br.readLine());
         } catch (IOException e) {
            e.printStackTrace();
            return "END_OF_FILE";
         }
      }
      return st.nextToken();
   }

   private int[] nextIntArr(int n) {
      int[] arr = new int[n];
      for (int i = 0; i < n; i++)
         arr[i] = nextInt();
      return arr;
   }

   private String[] nextArr(int n) {
      String[] arr = new String[n];
      for (int i = 0; i < n; i++)
         arr[i] = next();
      return arr;
   }
}