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 Python (3.8.2) (judge verdict: Accepted)

class int(int):  
  def times(self, callback):
    for times in range(0,self):
      callback()

class list(list):
  def maps(self, callback):
    return list(map(callback, self))

  def each(self, callback):
    list(map(callback, self))
    return self

  # index will sent to the function as a second parameter.
  def each_with_index(self,callback):
    list(map(callback, self, range(0,len(self))))
    return self

# Make sure you initialize object with class name.
# Ex: int(10).times, array = list([1,2,3,4,5])

target, length = input().split()
target = int(target)
length = int(length)
nums = list(input().split())

def convert(arg):
    return int(arg)
nums = nums.maps(convert)

for i in range(0,101):
  # Xの両隣がpに存在しないかみていく（pに存在しない値をみつけたいので）
  a = target - i # まずは左隣をみていく(Xよりi小さい)
  if a not in nums:
    print(a)
    exit()

  b = target + i # 次に右隣をみていく(Xより大きい)
  if b not in nums:
    print(b)
    exit()
  
# 配列pが与えられなかったとき(長さが0のとき)でも、「Xより0小さい値がpに存在しない」ので、そのままXが返せる