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>

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

  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (p[i] > p[j]) {
        tmp =  p[i];
        p[i] = p[j];
        p[j] = tmp;
      }
    }
  }

  ans = 0;
  for (i = 1; i < n; i++) {
    if (p[i] <= x) {
      if (p[i] != (p[i - 1] + 1)) {
        ans = p[i - 1] + 1;
//        printf("p[%d]: %d\n", i, ans);
      }
    }
    else if (p[i] > x) {
      if (p[i] != (p[i - 1] + 1)) {
        ans = p[i - 1] + 1;
//        printf("p[%d]: %d\n", i, ans);
      }
    }
  }
  
  printf("%d", ans);
  return (0);
}