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.ArrayList;
 
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int X = sc.nextInt();
      	int N = sc.nextInt();
      
      List <Integer> a = new ArrayList <Integer>();//リストでaを格納する
      
      for(int i=0; i<N; i++){
          a.add(sc.nextInt());//リストにaを入力していく,aはXに含まれない整数
      }
      
      for(int i=0; i<100; i++){//N(0~100)をまわす
      if(!a.contains(X-i)){//aに含まれないX-iの整数
         System.out.println(X-i);//Xに近い前の整数
         break;
      }
      else if(!a.contains(X+i)){
              System.out.println(X+i);//Xに近い後ろの整数
              break;
      }
      }
    }
}