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 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

public class Main{
//public class abc166_a{
	public static void main(String[] args){
		Scanner scan = new Scanner(System.in);
		
		int x = scan.nextInt();
		int n = scan.nextInt();
		int[] p = new int[101];
		for(int i=0; i<n; i++){
			p[scan.nextInt()] = 1;
		}
		
		int px = x;
		int mx = x;
		if(p[x] == 0)System.out.println(x);
		while(p[x] != 0){
			px += 1;
			mx -= 1;
			if(p[px] == 0){
				System.out.println(px);
				break;
			} 
			if(mx > 0){
				if(p[mx] == 0){
					System.out.println(mx);
					break;
				} 
			}else{
				//System.out.println(mx);
				//break;
			}
		}

	}

}