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: Wrong Answer)

import java.util.*;

class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    
    int x = sc.nextInt();
    int n = sc.nextInt();
    int[] p = new int[n];
    for(int i=0; i<n; i++){
      p[i] = sc.nextInt();
    }
    Arrays.sort(p);
    int[] np = new int[101-n];
    int a = 0;
    int b = 0;
    for(int i=0; i<=100; i++){
      if(a<n && p[a]==i){
        a++;
      }
      else{
        np[b++] = i-x;
      }
    }
    int min = 100;
    int ans = 0;
    for(int i=0; i<101-n; i++){
      if (min>Math.abs(np[i]) && ans > np[i]){
        min = Math.abs(np[i]);
        ans = np[i];
      }
    }
    System.out.println(ans+x);
  }
}