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.Scanner;

public class Main{

	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);

		String[] tmp = scanner.nextLine().split(" ");
		int x = Integer.parseInt(tmp[0]);
		int n = Integer.parseInt(tmp[1]);

		String[] tmp1 = scanner.nextLine().split(" ");
		int[] list = new int[102];
		int a = 0;
		list[0] = 0;
 		for(int i = 0; i < n; i++) {

			a = Integer.parseInt(tmp1[i]);

			list[a]++;

		}

		int ans = 101;
		int count = 101;
		for(int i = 101; i >= 0; i --) {


			if(list[i] != 1) {

				ans = Math.min(ans, Math.abs(x - i));

				if(ans == Math.abs(x - i)) {

					count = Math.min(count, i);

				}
			}

		}

		System.out.println(count);


	}

}