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.
#include <bits/stdc++.h>
#define STD /*
from sys import (
stdin, stdout, exit as sys_ret)
"""****************************

       Interactive Tasks:

          / Python: /       """
f_input, f_print, f_flush = (
        stdin.readline,
        stdout.write,
        stdout.flush)

"""        / C++ /
       #import <cstdio>
       fflush(stdout);
             or
       #import <iostream>
       cout << endl;

   —————————————————————————
    Don't raise your voice,
    improve your argument.
   —————————————————————————

   cat /dev/ass > /dev/head
            Ctrl+C
   cat /knowledge > /dev/head

                © Jakov Gellert
                        frvr.ru

****************************"""
# */ using namespace std; int

def the_good(amount, border, segments):
    intersection = [0 for i in range(border + 2)]
    for i in segments:
        intersection[i[0]] += 1
        intersection[i[1] + 1] -= 1
    for i in range(1, border + 1): intersection[i] += intersection[i - 1]
    ans = [i for i in range(1, border + 1) if intersection[i] == 0]
    f_print(str(len(ans)) + '\n')
    for i in ans:
        f_print(str(i) + ' ')

def the_bad(amount, border, segments):
    def bad(x): return False if len([i for i in segments if i[0] <= x and x <= i[1]]) else True
    ans = list(filter(bad, [i for i in range(1, border + 1)]))
    f_print(str(len(ans)) + '\n')
    f_print(' '.join([str(x) for x in ans]))

def the_ugly(amount, border, segments):
    ans = []
    for i in range(1, border + 1):
        flag = 0
        for j in segments:
            if j[0] <= i and i <= j[1]:
                flag = 1
                break
        if flag == 0:
            ans.append(i)
    f_print(str(len(ans)) + '\n')
    for i in ans:
        f_print(str(i) + ' ')
    
amount, border = map(int, f_input().split())
segments = [[int(__) for __ in f_input().split()] for _ in range(amount)]

the_good(amount, border, segments)
#the_bad(amount, border, segments)
# and (task: 1015A)
#the_ugly(amount, border, segments)