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 1.8.0) (judge verdict: Wrong Answer)

import java.util.HashSet;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int integerX = sc.nextInt();  //元の整数
		int youso = sc.nextInt();  //整数列の要素数
		HashSet<Integer> list = new HashSet<Integer>();  //整数列を入れるリスト

		//配列を入力
		for(int i=0 ; i<youso ; i++) {
			list.add(sc.nextInt());
		}
        
		//配列の要素数が0なら0と表示して終了
		if(list.isEmpty() == true) {
			System.out.println(0);
			return;
		}
        
        //指定した整数を+1、-1ずつする
		int i = integerX-1;
		int j = integerX+1;

		//iに+1、jに-1をしながら、配列に含まれるのか判断
		for(;;) {
			if(list.contains(i)) {
				i--;
			}
			else{
				System.out.println(i);
				return;
			}

			if(list.contains(j)) {
				j++;
			}
			else {
				System.out.println(j);
				return;
			}
		}
	}
}