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.
from sys  import stdin,stdout
 
import bisect

import math

from collections import Counter
 
def st():
    return list(stdin.readline().strip())
 
def inp():
    return int(stdin.readline())
 
def li():
    return list(map(int,stdin.readline().split()))
 
def mp():
    return map(int,stdin.readline().split())
 
def pr(n):
    stdout.write(str(n)+"\n")
 
def soe(limit):
    l=[1]*(limit+1)
    prime=[]
    for i in range(2,limit+1):
        if l[i]:
            for j in range(i*i,limit+1,i):
                l[j]=0
 
    for i in range(2,limit+1):
        if l[i]:
            prime.append(i)
    return prime
 
def segsoe(low,high):
    limit=int(high**0.5)+1
    prime=soe(limit)
    n=high-low+1
    l=[0]*(n+1)
    for i in range(len(prime)):
        lowlimit=(low//prime[i])*prime[i]
        if lowlimit<low:
            lowlimit+=prime[i]
        if lowlimit==prime[i]:
            lowlimit+=prime[i]
        for j in range(lowlimit,high+1,prime[i]):
            l[j-low]=1
    for i in range(low,high+1):
        if not l[i-low]:
            if i!=1:
                print(i)
 
def power(a,n):
    r=1
    while n:
        if n&1:
            r=(r*a)
            n=n-1
        else:
            a=(a*a)
            n=n>>1
    return r
    
def solve():
    n,m=mp()
    l=[0]*m
    for i in range(n):
        a,b=mp()
        for i in range(a-1,b):
            l[i]=1
    pr(l.count(0))
    for i  in range(m):
        if l[i]==0:
            print(i+1,end=" ")
            
            

for _ in range(1):
    solve()
