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: Runtime Error)

import java.util.*;
public class Main  {
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);

      int x = sc.nextInt();
      int n = sc.nextInt();
      
      if (n == 0) {
         System.out.println(x);
      } else {
         int a[] = new int[n];

         for (int i = 0; i < n; i++) {
            int aa = sc.nextInt();
            a[i] = aa;
         }

         Arrays.sort(a);
      
         int last = a[n-1];
      
         int b[] = new int[last];
      
         for (int j = 0; j < n; j++) {
            b[j] = 0;
         }
      
         for (int k = 0; k < n; k++) {
            int aaa = a[k];
            b[aaa-1] = 1;
         }
      
         if (b[x] == 0) {
            System.out.println(x);
         } else {
            int fin = 0;
            int count = 1;
            while (fin == 0) {
               if (b[x-count-1] == 0) {
                  System.out.println(x-count);
                  fin++;
               } else if (fin == 0 && last >= x + count &&  b[x + count-1] == 0) {
                  System.out.println(x+count);
                  fin++;
               }
               count++;
            }
         }
      }
   }
}