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;
using ll=long long;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define inf 1001001001
#define mod 1000000007
#define mod2 998244353
#define pi acos(-1)
#define all(v) v.begin(),v.end()

ll gcd(ll a,ll b){
    if(b>a) swap(a,b);
    if(b==0) return a;
    return gcd(b,a%b);
}
ll lcm(ll a,ll b){
    ll g=gcd(a,b);
    return a/g*b;
}
ll rep_jijo(ll n,ll x){
    if(x==0) return 1;
    if(x%2==0){
        ll t=rep_jijo(n,x/2);
        return t*t;
    }
    return n*rep_jijo(n,x-1);
}

int main(){
    int x,n;
    cin >> x >> n;
    int ans = inf;
    int ansp;
    vector<int>p(n);
    vector<bool>num(101,true);
    rep(i,n){
        int a;
        cin >> a;
        num[a] = false;
    }
    for(int i = 1; i<=100; i++){
        if(num[i]){
            int sub = abs(x-i);
            if(sub < ans){
                ans = sub;
                ansp = i;
            }
        }
    }
    if(abs(0-x) < ans){
        ansp = 0;
    }
    if (abs(101-x) < ans ){
        ansp = 101;
    }
    cout << ansp <<endl;
   return 0;
}