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<math.h>
#include<string.h>
int main()
{
    long long int ch[100000];
    char arr[1000000];
    char kh,bh,ag[100];
    int cp[400000]= {0};
    int a,b,c,d,e=0,g,i=0,tep=0,temp=0,j=0,k=0,l=0,m=0,n=0;
    scanf("%d %d",&a,&b);
    i=100;
    for(k=1;k<=b;k++)
    {
     scanf("%d",&c);
     cp[c]=1;
     if(i>c)
     i=c;
    }
    j=a;
    k=100;
    l=0;
    for(i;i<=100;i++)
    {
     if(cp[i]==0)
     {
      if(abs(j-i)<k)
       {
        k=abs(j-i);
        l=i;
       }
     }
    }
    printf("%d\n",l);
    return 0;
}