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.*;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		final int X = sc.nextInt();
		final int N = sc.nextInt();
		if(N==0){
			sc.close();
			System.out.println(X);
			return;
		}
		ArrayList<Integer> p = new ArrayList<>();
		int max=0;
		for(int i=0;i<N;++i){
			p.add(sc.nextInt());
		}
		for(int i=0;i<N;++i){
			max = Math.max(p.get(i), max);
		}
		sc.close();
		int ans = 0;
		int i=0;
		int min = 101;
		int tmp = min;
		while(i<(X+max)){
			if(p.contains(i)) {
				++i;
				continue;
			}
			min = Math.min(Math.abs(X-i), min);
			if(Math.min(tmp, min)==tmp){break;}
			ans = i;
			++i;
			tmp = min;
		}
		System.out.println(ans);
	}
}