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 <string.h>
#include <stdbool.h>
#include <math.h>

int main(void) {
    int X, N;
    int p[100];

    for(int i = 0; i < 100; i++) {
        p[i] = 0;
    }

    scanf("%d %d", &X, &N);
    for(int i = 0; i < N; i++) {
        int tmp;
        scanf("%d", &tmp);
        p[tmp-1] = 1;
    }

    for(int i = 0; i <= 100; i++) {
        if(p[X-i-1] == 0) {
            printf("%d\n", X-i);
            return 0;
        }
        if(p[X+i-1] == 0) {
            printf("%d\n", X+i);
            return 0;
        }
    }

    return 0;
}