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.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 a[]=new int[n];
		for(int i=0;i<n;i++) {
			a[i]=sc.nextInt();
		}
		if(n==0) {
			System.out.println(x);

		}else {
			Arrays.sort(a);

			int j=0;
			int p=100;
			int q=100;
			for(j=0;a[j]<x;j++) {

			}
			int c=0;
			int i=0;
			boolean f=true;
			for(i=0;i<=x+100;i++) {
				if(x-i!=a[Math.max(0, j-c)]) {

					break;
				}else if(x+i!=a[Math.min(n, j+c)]) {
					f=false;
				    break;
				}else {

					    c++;
				}
			}

			if(f) {
				System.out.println(x-i);
			}else {
				System.out.println(x+i);
			}
		}





	}

}