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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n=scan.nextInt();
		int x=scan.nextInt();
		int [] a=new int[x];
		ArrayList a1=new ArrayList();
		ArrayList a2=new ArrayList();
		for(int i=0;i<x;i++)
		{
			a[i]=scan.nextInt();
			a1.add(a[i]);
		}
		Arrays.sort(a);
		int b=n;
		
		while(a.length>0&&b>a[0])
		{
			
			b--;
			if(a1.contains(b))
			{
				continue;
			}
			else
			{
				a2.add(b);
				break;
			}
		}
		while(true)
		{
			b++;
			if(a1.contains(b))
			{
				continue;
			}
			else
			{
				a2.add(b);
				break;
			}
		}
		if(a.length==0) System.out.println(n);
		else System.out.println(Collections.min(a2));
		
	}

}