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: Accepted)

#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <stdio.h>
int compare(const void *a,const void *b){
    if(*(long long *)a>*(long long *)b)return 1;
    return -1;
}
long long array[1000001]={0};
int main(int argc, const char * argv[]) {
    int n,x;
    scanf("%d%d",&x,&n);
    long long p[n];
    for(int i=0;i<n;i++){
        scanf("%lld",&p[i]);
    }
    //qsort(p,n,sizeof(long long),compare);
    int array[105];
    for(int i=0;i<=102;i++){
        array[i]=1;
    }
    for(int i=0;i<n;i++){
        array[p[i]]=0;
    }
    int d=200,ans=-1;
    for(int i=-100;i<=101;i++){
        if(abs(x-i)<d&&array[i]){
            ans=i;
            d=abs(x-i);
        }
    }
    printf("%d",ans);
    
    
    return 0;
}