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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int X = sc.nextInt();
		int N = sc.nextInt();
		int min = 0;
		int max = 101;
		Integer[] p = new Integer[N];
		for(int i=0; i<N ;i++) {
			p[i] = sc.nextInt();
		}
	if(N != 0) {
		for(int i = 1; i <=101; i++) {
			//配列内に要素が存在していなければ入る
			if(!Arrays.asList(p).contains(i)) {
				min = min<=i && i<=X? i:min;
				max = i<=max && i>=X? i:max;
			}
		}
		if(Math.abs(min-X) <= Math.abs(max-X)){
			System.out.println(min);
		}else {
			System.out.println(max);
		}

	}else {
		System.out.println(X);
	}

	}
}