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)

/*
    forbidden_list.c
    2020/06/14
*/

#include <stdio.h>

int main()
{
    int x, n, i, j, tmp;
    int arr[102] = {0};
    scanf("%d %d", &x, &n);
    if (n == 0){
        printf("%d", x);
        return 0;
    }
    if (n == 100){
        printf("%d", x);
        return 0;
    }
    for (i = 0; i < n; i++){
        scanf("%d", &tmp);
        arr[tmp] = 1;
    }
    for (j = 0; (0 <= x - j)&&(x + j <= 100); j++){
        if(arr[x - j] == 0){
            printf("%d", x - j);
            break;
        }
        if(arr[x + j] == 0){
            printf("%d", x + j);
            break;
        }
        
    }
   return 0;
}