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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;} 
int main(void)
{
    int x,n,p[100],ans=0,a[100],b;
    scanf("%d %d",&x,&n);
    for(int i=0;i<n;i++)
        scanf("%d",&p[i]);

    for(int i=0;i<100;i++)
        a[i]=i;

    for(int i=0;i<n;i++)
        a[p[i]]=0;

    for(int i=-100;i<101;i++)
    {
        b=abs(x-i);
        if(b<abs(x-ans)&&a[i])
            ans=i;
    }
printf("%d",ans);

}