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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
import java.util.Arrays;
 
public class Main{
  public static void main(String[] args){
    
    Scanner sc = new Scanner(System.in);
    int x = sc.nextInt();
    int n = sc.nextInt();
    ArrayList<Integer> list = new ArrayList<>();
    for(int i = 0; i < n; i++){
         list.add(sc.nextInt());
    }
    
    if(n == 0){
      System.out.println(x);
      return;
    }else if(!list.contains(x)){
      System.out.println(x);
      return;
    }else{
 
      int i = 1;
      while(true){
        int a = x-i;
        int b = x+i;
        if(!list.contains(a)){
          System.out.println(a);
          return;
        }else if(!list.contains(b)){
          System.out.println(b);
          return;
        }else{
          i++;
        } 
      }
      
    }
    
  }
}