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.*;
import java.math.RoundingMode;
import java.math.BigDecimal;
 
 
 
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];
		for(int i = 0; i < n; i++) {
			p[i] = sc.nextInt();
		}
		Arrays.sort(p);
		int ans = x;
		int sub = 100;
		if(n==0) ans = x;
		else if(x<p[0]) ans=x;
		else if(x>p[n-1]) ans=x;
		else {
			if(x==p[0]) ans=x-1;
			else if(p[n-1]==x&&p[n-1]-p[n-2]<=1) ans=x+1;
			else {
				for(int i = 0; i < n-1; i++) {
					int left = p[i];
					int right = p[i+1];
						if(left>=x&&sub>(left+1)-x) {
							if(right-left>1) {
								ans=left+1;
								sub = ans-x;
							}
							
						}
						else if(right<=x&&sub>x-(right-1)) {
							if(right-left>1) {
								ans=right-1;
								sub = x-ans;
							}
							
						}
						
				}
			}
		}
		
		System.out.println(ans);
	 }
}