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<bits/stdc++.h>
using namespace std;
int main(){
    int x,n;
    cin>>x>>n;
    vector<int>minas;
    vector<int>plus;
    bool zero=true;
    for(int i=0;i<n;i++){
        int current;
        cin>>current;
        if(current-x<0)
        {
            minas.push_back(current-x);
        }
        else if(current-x>0)
        {
            plus.push_back(current-x);
        }
        else{
            zero=false;
        }
        
    }
    sort(minas.begin(),minas.end());
    reverse(minas.begin(),minas.end());
    sort(plus.begin(),plus.end());
 
    if(zero)
    {
        cout<<x<<endl;
    }
    else if(minas.size()==0)
    {
        for(int i=1;true;i++)
        {
            if(i>plus.size()||i!=plus.at(i-1))
            {  
                cout<<x+i<<endl;
                break;
            }
        }
    }
    else if(plus.size()==0)
    {
        for(int i=1;true;i++)
        {
             if(i>minas.size()||-i!=minas.at(i-1))
            {  
                cout<<x-i<<endl;
                break;
            }
        }
    }else
    {
         for(int i=1;true;i++)
        {
             if(i==minas.size()+1||-i!=minas.at(i-1))
            {  
                cout<<x-i<<endl;
                break;
            }
            if(i==plus.size()+1||i!=plus.at(i-1))
            {  
                cout<<x+i<<endl;
                break;
            }
        }
    }
}