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 <stdlib.h>

int main() {
  int x, count;
  scanf("%d%d", &x, &count);
  int array[count];
  for (int i = 0; i < count; i++) {
    scanf("%d", &array[i]);
  }
  int array2[100 - count];
  int check = 0;
  for (int j = 0; j <= 100 - count; j++) {
    for (int k = 0; k < count; k++) {
      if (j != array[k]) {
        check++;
      }
    }
    if (check == count) {
      array2[j] = j;
    }
    check = 0;
  }
  int tmp = abs(x);
  int tmp2 = 0;
  int index = 0;
  for (int l = 0; l <= 100 - count; l++) {
    tmp2 = abs(x - array2[l]);
    if (tmp > tmp2) {
      tmp = tmp2;
      index = l;
    }
  }
  printf("%d\n", index);
  return 0;
}