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>

int main(){
int x=0;
int p[100]={};
int n=0;
int dist=101;
int ans=0;
int z[101]={};
for(int i=0;i<101;i++){
    z[i]=i;
}
scanf("%d %d",&x,&n);

if(n!=0){
for(int i=0;i<n;i++){
    scanf("%d",&p[i]);
z[p[i]]=10000;
}
}
if(n==100){
    if(x<51){
        ans=0;
    }else{
        ans=101;
    }

}else{
for(int i=0;i<101;i++){
    if(dist>=abs(x-z[i])){
        if(dist==abs(x-z[i])){
            if(ans<z[i]){
                z[i]=ans;
            }
        }
        dist=abs(x-z[i]);
        ans=z[i];
    }
}
}

printf("%d\n",ans);

return 0;

}