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: Wrong Answer)


import java.util.*;

public class Main {
    public static void main(final String[] args) {
        final Scanner sc = new Scanner(System.in);
        // 整数の入力
        // final int a = sc.nextInt();
        // 文字列の入力
        // final String s1 = sc.next();
        final double x = sc.nextDouble();
        final int n = sc.nextInt();
        List<Double> ngList = new ArrayList<Double>();
        for (int i = 0; i < n; i++) {
            ngList.add(sc.nextDouble());
        }
        sc.close();
        double diff = 100;
        int result = 0;
        for (int i = 1; i <= 100; i++) {
            if (diff > Math.abs(x - i) && !ngList.contains((double) i)) {
                diff = Math.abs(x - i);
                result = i;
            }

        }
        System.out.println(result);
    }
}