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.io.*;

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

      int x = sc.nextInt();
      int n = sc.nextInt();
      int[] arr = new int[n];
      for(int i = 0; i < n; i++){
          arr[i] = sc.nextInt();
      }
      Arrays.sort(arr);
      
      int best = x;
      int minDistance = n;
      for(int i = 0; i <= 101; i++){
          boolean good = true;
          for(int j = 0; j < n; j++){
              if(arr[j] == i){
                  good = false;
                  break;
              }
          }
          if(good){
              if(Math.abs(x - i) < minDistance){
                  minDistance = Math.abs(x-i);
                  best = i;
              } else if(Math.abs(x-i) == minDistance){
                  best = Math.min(best, i);
              }
          }
      }
      System.out.println(best);
    }
}