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.Scanner;

public class Main {

	public static void main(String[] args) throws Exception{
		Scanner scanner = new Scanner(System.in);
		int X = Integer.parseInt(scanner.next());
		int N = Integer.parseInt(scanner.next());

		ArrayList<Integer> list = new ArrayList<Integer>();

		for(int i=0;i<N;i++) {
			list.add(Integer.parseInt(scanner.next()));
		}
		int count = 1;
		if(N==0) {
			System.out.print(X);
		}else {
			while(true) {
				if(!list.contains(X-count)) {
					System.out.print(X-count);
					break;
				}else if(!list.contains(X+count)) {
					System.out.print(X+count);
					break;
				}
				count++;
			}
		}

		scanner.close();
	}
}