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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
//#define f(i,a,n) for(ll i=a;i<n;i++)
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)


 
int main() 
{
    fast;
    ll t=1;
    //cin>>t;
    while(t--)
    {
        ll x,n,p,q,miin=9999999999,k=0;
        cin>>x>>n;
        ll A[n],B[1000]={0},ans=99999999999;
        if(n==0)
        cout<<x;
        else
        {
        for(ll i=0;i<n;i++)
        {
            cin>>A[i];
            B[A[i]]++;
            
        }
        for(ll i=0;i<=101;i++)
        {
            if(B[i]==0)
            {
                
                //cout<<i<<" ";
                p=abs(x-i);
                //cout<<p<<" ";
                if(p<=miin)
                {
                    if(p==miin)
                    {
                    miin=p;
                    ans=min(i,ans);
                    //cout<<ans<<" ";
                    }
                    else
                    {
                        miin=p;
                        ans=i;
                    }
                }
            }
            
        }
        cout<<ans;
        }
    }
    return 0;
}