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: Accepted)

import java.util.Scanner;

public class Main
{
	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt();
		int b = sc.nextInt();
		if (b==0) {
			System.out.println(a);
			return;
		}
		int [] c = new int[b];
		String str = sc.nextLine();
		int min = 100;
		int ans = 0;
		for (int i=0; i<b; i++) {
			c[i]= sc.nextInt();
		}
		int f = 0;
		for (int i=-101; i<201; i++) {
			for (int j=0; j<b; j++) {
				if (c[j]==i) {
					f = 1;
					break;
				}
			}
			if (f == 0) {
				if(Math.abs(i-a) < min) {
					min = Math.abs(i-a);
					ans = i;
				}
			} else {
				f = 0;
			}
		}

		sc.close();
		System.out.println(ans);
	}
}