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(){
    int x, n, p[101], i, j, a[101];
    scanf("%d %d", &x, &n);
    for(i=0;i<n;i++){
        scanf("%d", &p[i]);
    }
    if(n == 0){
        printf("%d", x);
    }else{
        for(j=1;j<=100;j++){
            for(i=0;i<n;i++){
                if(j == p[i]){
                    a[j] = 10000;
                }else{
                    a[j] = fabs(x - j);
                }
            }
        }
        j = 0;
        int min=a[j];
        for(j=0;j<100;j++){
            if(min > a[j]) min=a[j];
        }
        printf("%d",x-min);
    }
    return 0;
}