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 <stdio.h>
signed main(){
    int i,x,n,p[105];
    scanf("%d %d",&x,&n);
    for(i=0;i<n;i++){
        scanf("%d",&p[i]);
    }
    int flag[105]={};
    int ans=0;
    for(i=0;i<n;i++){
        flag[p[i]]=1;
    }
    for(i=1;i<=101;i++){
        if(flag[i]==0){
            int abs1,abs2;
            abs1=x-ans;
            abs2=x-i;
            if(abs1<0){
                abs1=abs1*-1;
            }
            if(abs2<0){
                abs2=abs2*-1;
            }
            if(abs1>abs2){
                ans=i;
            }
        }
    }
    printf("%d\n",ans);
    return(0);
}