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.List;
import java.util.Scanner;
import java.util.ArrayList;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int x = sc.nextInt();
		int n = sc.nextInt();
		int p[] = new int[n];
		List<Integer> all = new ArrayList<>();
		for(int i=0; i<n; i++) {
			p[i] = sc.nextInt();
		}
		int flag=0;
		for(int i=1; i<=100; i++) {
			flag=0;
			for(int j=0; j<n; j++) {
				if(p[j]==i) {
					flag=1;
				}
			}
			if(flag==0) {
				all.add(i);
			}
		}

		
		int min = 100;
		int ans = x;
		for(int i=0; i<all.size(); i++) {
			if(min > Math.abs(x - all.get(i))) {;
			min = x-all.get(i);
			ans = all.get(i);
			}
		}
		System.out.println(ans);
	}
}