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 int long long
#define INT_MAX INT64_MAX
#define INT_MIN INT64_MIN
#define PI 3.14159265358979323846
#define CLK clock_t clk = clock();//Start of main
#define OCLK cerr << "Time (in ms): " << (double)(clock() - clk) * 1000.0 / CLOCKS_PER_SEC << '\n';//End of main
#define MOD 1000000007
/* DEFINE cout */
#define pln cout << "===============================\n"
#define on cout<<"\n"
#define os cout<<" "
#define o2(a,b) cout<<a<<" "<<b
#define o(a) cout << a
#define bitcount __builtin_popcount
/* DEFINE cout */
#define endl "\n"
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);//fastio
using namespace std;
struct hash_pair { // use pair as key in unordered_map<pair<int,int>,int,hash_pair>;
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2>& p) const
    {
        auto hash1 = hash<T1>{}(p.first);
        auto hash2 = hash<T2>{}(p.second);
        return hash1 ^ hash2;
    }
};
bool prime[10000001];
int P=MOD-2;
int factorial[10000001]={0};
int powerFunction(int x,int y){
    int res = 1;
    int p = MOD;
    x = x % p;
    while (y > 0){
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}
void sieveFunction(int maxLimit){
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1]=false;
    for(int i=2; maxLimit >= i ; i++){
        if(prime[i]){
            for(int j=2*i ; maxLimit >= j ; j+=i)
                prime[j]=false;
        }
    }
}
void factorialFunction(int maxLimit){
    factorial[0]=1;
    for(int i=1 ; i <= maxLimit ; i++)
        factorial[i]=(factorial[i-1]*i)%MOD;
    return;
}
int gcd(int a, int b){
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
signed main(){
    int x, n;
    cin >> x >> n;
    map<int, bool> mp;
    for(int i=0; n>i; i++){
        int num;
        cin >> num;
        mp[num] = true;
    }
    for(int i=0; 200>i; i++){
        if(mp[x-i] == false){
            cout << x-i << endl;
            return 0;
        }
        if(mp[x+i] == false){
            cout << x+i<< endl;
            return 0;
        }
    }
}