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>

int main(void){
    int x,n=0;
    int temp=0;
    int flagP=0;
    int flagM=0;
    int array[100]={0};

    scanf("%d %d",&x,&n);
    if(n==0){
        printf("%d",x);
        return 0;
    }

    for(int i=0;i<n;i++){
        scanf("%d",&array[i]);
    }


    for(int i=0;i<n;i++){
        flagP=0;
        flagM=0;
        for(int j=0;j<n;j++){
            if(array[j]==x+i){
                flagP=1;
            }
            if(array[j]==x-i){
                flagM=1;
            }
            if(flagP==1 && flagM==1){
                break;
            }
        }
        if(flagP==0 || flagM==0){
            temp=i;
            break;
        }
    }
    if(flagM==0){
        printf("%d",x-temp);
    }
    else if(flagP==0){
        printf("%d",x+temp);
    }


    return 0;
}