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>
#define ll long long int
using namespace std;


int main () {
ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll x;
    cin>>x;
    ll n;
    cin>>n;
    ll a[n];
    if(n==0)
    cout<<x<<endl;
    ll i,j;
    for(i=0;i<n;i++)
    {
    	cin>>a[i];
	}
    if(n==1)
    {
    	if(x<=a[0])
    	cout<<a[0]-1<<endl;
    	else
    	cout<<a[0]+1<<endl;
	}
	
	if(n>1)
	{
		vector<ll>k;
		k.clear();
		sort(a,a+n);
		for(i=1;i<n;i++)
		{
			ll z=a[i]-a[i-1];
			if(z==1)
			continue;
			else
			{
				for(j=a[i-1]+1;j<a[i];j++)
				{
					//cout<<"j ="<<j<<endl;
					k.push_back(j);
				}
			}
		}
		ll v=k.size();
		//cout<<"V="<<v<<endl;
		ll diff;ll ans=INT_MAX;diff=INT_MAX;
		//cout<<"OK"<<endl;
		for(i=0;i<v;i++)
		{
			ll s=abs(x-k[i]);
			if(s<diff)
			{
				//cout<<"Hm"<<endl;
				diff=s;
				ans=k[i];
			}
		}
		ll q=abs(x-(a[0]-1));
		if(q<diff)
		{
			ans=a[0]-1;
			diff=q;
		}
		ll w=abs(x-(a[n-1]+1));
		
		if(w<diff)
		{
			ans=a[n-1]+1;
			diff=w;
		}	
		
		cout<<ans<<endl;
		
	}
	return 0;
}