Problem:
Score : 300 points

Problem Statement

Given are an integer X and an integer sequence of length N: p_1, \ldots, p_N.

Among the integers not contained in the sequence p_1, \ldots, p_N (not necessarily positive), find the integer nearest to X, that is, find the integer whose absolute difference with X is the minimum. If there are multiple such integers, report the smallest such integer.

Constraints

1 \leq X \leq 100

0 \leq N \leq 100

1 \leq p_i \leq 100

p_1, \ldots, p_N are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X N
p_1 ... p_N

Output

Print the answer.

Sample Input 1

6 5
4 7 10 6 5

Sample Output 1

8

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the one nearest to 6 is 8.

Sample Input 2

10 5
4 7 10 6 5

Sample Output 2

9

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the ones nearest to 10 are 9 and 11. We should print the smaller one, 9.

Sample Input 3

100 0

Sample Output 3

100

When N = 0, the second line in the input will be empty. Also, as seen here, X itself can be the answer.

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

#ABC170C.rb
x,n = gets.split.map(&:to_i)
#ar = gets.split.map{|n| n.to_i.abs}
ar = gets.split.map(&:to_i)
ar_max = ar.max
if n == 0
  puts x
  exit
elsif x >= ar_max
  br = (1..x).to_a
else
  br = (1..ar_max).to_a
end

br.delete_if do |br2|
  ar.include?(br2)
end

##########
if br.empty?
  puts x
  exit
end


min_diff = (x-br[0]).abs
min = br[0]
if n == 1
  puts min
  exit
end


br[1..br.size].each do |br3|
  diff = (x-br3).abs
  if diff < min_diff
    min_diff = diff
    min = br3
    #puts "#{br3}:min_diff=#{min_diff} min=#{min}"
  else
     break
  end
end

puts min