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.Scanner;
public class A {
    private int[] l,r;
    private int n,m;
    public A(){
        l = new int[110];
        r = new int[110];
        n = 0;
        m = 0;
    }
    public void input(){
        Scanner sc = new Scanner(System.in);
        this.n = sc.nextInt();
        this.m = sc.nextInt();
        for(int i=0;i<n;i++){
            this.l[i] = sc.nextInt();
            this.r[i] = sc.nextInt();
        }
    }

    public void solve(){
        int res =0;
        boolean[] free = new boolean[110];
        for(int i =0;i<110;i++) free[i] = false;
        for(int i=1;i<=m;i++){
            boolean ok = true;
            for(int j=0;j<n;j++) if (l[j]<=i && r[j] >=i) {ok = false; break;}
            if (ok) {
                res++;
                free[i] = true;
            }
        }
        System.out.println(res);
        for(int i=0;i<110;i++) if (free[i]) System.out.print(i + " ");
    }
    public static void main(String[] args){
        A duc = new A();
        duc.input();
        duc.solve();
    }
}