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 <stdlib.h>
#include <bits/stdc++.h>
#include <math.h>
using namespace std;
using ll=long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
inline constexpr ll gcd(ll a,ll b){if(!a||!b)return 0;while(b){ll c=b;b=a%b;a=c;}return a;}
inline constexpr ll lcm(ll a,ll b){if(!a||!b)return 0;return a*b/gcd(a,b);}
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
const long long INF=1LL << 60;
long long a,b;
long long dp[100200][110];
long long w[110],v[110];

static const double pi = 3.141592653589793;
int main(){
    int a,b;
    cin>>a>>b;
    bool f=true;
    vector<int> vec(b);
    rep(i,b){
        cin>>vec.at(i);
        if(vec.at(i)==a){
            f=false;
        }
    }
    if(f){
        cout<<a<<endl;
    }
    else{
        rep(i,101){
            bool f2=true;
            bool f3=true;
            rep(j,b){
                if((a-(i+1))==vec.at(j)){
                    f2=false;
                }
                if((a+(i+1))==vec.at(j)){
                    f3=false;
                }
            }
            if(f2){
                cout<<a-(i+1)<<endl;
                break;
            }
            if(f3){
                cout<<a+(i+1)<<endl;
                break;
            }
        }
    }
    
}