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)

//package atcoder;
 
import java.util.*;
//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();
		Integer p[] = new Integer [N];
		for(int i=0;i<N;i++) {
			p[i] = sc.nextInt();
		}
		/*for(int i=0;i<N;i++) {
			System.out.println(p[i]);
		}*/
	    int min=1000;
		for(int j=0;j<101;j++) {
			if(Arrays.asList(p).contains(j)) {
				//System.out.println(j);
				continue;
			}
			//System.out.println(j);
		  min = Math.min(min, Math.abs(X-j));
		}
		int a = X-min;
		if(Arrays.asList(p).contains(a)) {
			System.out.println(X+min);
		}else {
			System.out.println(a);
		}
	}
}