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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#define N 101

int main()
{
    int x, n, p[N], check[N+1], min, sub = -1;
    for (int i = 0; i < n; i++)
        check[i] = 0;

    scanf("%d %d", &x, &n);
    for (int i = 0; i < n; i++)
        scanf("%d", &p[i]);
    if(n==0){
        printf("%d",x);
        return 0;
    }

    for (int i = 0; i < n; i++)
        check[p[i]] = 1;

    for (int i = 0; i <= N; i++)
    {
        if (check[i] == 0)
        {
            if (abs(x - i) < sub || sub == -1)
            {
                min = i;
                sub = abs(x - i);
            }
        }
    }

    printf("%d", min);
    return 0;
}