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 pfi(a) printf("%d",a);
#define pfl(a) printf("%lld ",a);
#define pfin(a) printf("%d\n",a);
#define pfln(a) printf("%lld\n",a);
#define sfi(a) scanf("%d",&a);
#define sfl(a) scanf("%lld",&a);
#define pb push_back
#define f first
#define s second
#define vi vector<int>
#define vl vector<long long>
#define vs vector<string>
#define watch(x) cout << (#x) << " is " << (x) << endl
#define d0(x) cout<<(x)<<" "
#define d1(x) cout<<(x)<<endl
#define d2(x,y) cout<<(x)<<" "<<(y)<<endl
#define d3(x,y,z) cout<<(x)<<" "<<(y)<<" "<<(z)<<endl
#define d4(a,b,c,d) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<endl
#define d5(a,b,c,d,e) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<" "<<(e)<<endl
#define fo(i,a,b) for(int i=a;i<b;i++)
#define all(c) (c).begin(),(c).end()
#define ll long long
const int M=1e9+7;
#define endl "\n"
const double pyi=3.14159265358979323846;

int main() 
{ 
    int a,n;
    cin>>a>>n;

    int arr[n];
    unordered_set<int> stt;
    for(int i=0;i<n;i++)
    {
    	cin>>arr[i];
    	stt.insert(arr[i]);
    }
    
    
    int diff=1e7;
    int ans;
    for(int i=-105;i<105;i++)
    {
    	if(stt.find(i)!=stt.end()) continue;
        if(abs(i-a)<diff)
        {
        	diff=abs(i-a);
        	ans=i;
        }
    }
    cout<<ans<<endl;
    return(0);
}