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 <vector>
#include <algorithm>
using namespace std;

int main() {

int x,n, r{101};
vector<int>v{};
cin>>x>>n;
if(!n)
  cout << x;
else
{

  int c;
  for(int i = 0; i < n; i++)
  {
    cin >> c;
    v.push_back(abs(c));
  }

  sort(v.begin(), v.end());
  auto it = find(v.begin(),v.end(), x);
  
  if (it == v.end()){
    cout << x;
    exit(0);
  }
    
 
  
  int z = 1;
  while(true)
  {
    if(*(it-z) != x-z)
    {
      z = x-z;
      break;
    }
    else if (*(it+z) != x+z)
    {
      z = x+z;
      break;
    }
    z++;
    
  }
  cout << z;
  



}
return 0;
}