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: Runtime Error)

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 hako[] = new int[101];
		for(int i=0;i<101;i++) {
			hako[i]=0;
		}
		for(int i=0;i<n;i++) {
			int a = sc.nextInt();
			hako[a] = 1;
		}
		
		if(hako[x]==0) {
			System.out.println(x);
			return;
		}
		int up=x+1;
		int down=x-1;
		
		while(up>-1 && down<101) {
			if(hako[down] == 0) {
				System.out.println(down);
				return;
			}
			if(hako[up] == 0) {
				System.out.println(up);
				return;
			}
			up++;
			down--;
		}
		while(down<101) {
			if(hako[down] == 0) {
				System.out.println(down);
				return;
			}
			down--;
		}
		while(up>-1) {
			if(hako[up] == 0) {
				System.out.println(up);
				return;
			}
			up++;
		}
		
		
	}
}