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 <iostream>
#include <string>
#include <algorithm>
#include <math.h>
#include <vector>

using namespace std;

int main()
{
  int x, n;
  cin >> x >> n;
  if (n == 0)
  {
    cout << x << "\n";
    return 0;
  }

  int v[101];
  for (int i = 0; i <= 100; i++)
  {
    v[i] = 0;
  }

  int p;
  for (int i = 0; i < n; i++)
  {
    cin >> p;
    v[p] = 1;
  }

  int res = x;
  int i = 1;
  while (true)
  {
    if (x - i <= 0)
    {
      res = x - i;
      break;
    }
    if (x + i >= 101)
    {
      res = x + i;
      break;
    }

    if (v[x - i] == 0)
    {
      res = x - i;
      break;
    }
    if (v[x + i] == 0)
    {
      res = x + i;
      break;
    }

    i++;
  }

  cout << res << "\n";
  return 0;
}