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;

typedef long long ll;
typedef long double ld;

#define pb push_back
#define sort(s) sort(s.begin(),s.end())
#define reverse(s) reverse(s.begin(),s.end())
#define rep(i,n) for(ll (i)=0;(i)<(n);(i)++)

const ll mod = 1e9+7;

//最大公約数
ll gcd(ll a, ll b) {
    return b ? gcd(b, a%b) : a;
}

//最小公倍数
ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}

//素数判定
bool isPrime(ll x){
    ll i;
    if(x < 2)return 0;
    else if(x == 2) return 1;
    if(x%2 == 0) return 0;
    for(i = 3; i*i <= x; i += 2) if(x%i == 0) return 0;
    return 1;
}

//桁和
int digsum(ll n) {
    int res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

//桁数
int dignum(ll n) {
    int res = 0;
    while(n > 0) {
        res++;
        n /= 10;
    }
    return res;
}

//文字列中の特定の文字カウント
ll stringcount(string s,char c){
    return count(s.cbegin(),s.cend(),c);
}

//階乗
ll ka(ll i){
    ll res=1;
    while(i>0){
        res=res*i;
        i--;
    }
    return res;
}
 
//ncr
ll ncr(ll x,ll y){
    ll a = x;
    ll b = 1;
    if(y>(x/2)){
        y=x-y;
    }
    for(ll i=1;i<y;i++){
        a*=x-i;
        b*=i+1;
    }
    return a/b;
}

//フィボナッチ数列
ll f(ll x){
    ll dp[x+1];
    dp[1]=1;
    dp[2]=1;
    for(ll i=3;i<=x;i++){
        dp[i]=dp[i-1]+dp[i-2];
    }
    return dp[x];
}

//char->int
int ctoi(char c){
    return c-'0';
}

vector<pair<long long, long long> > prime_factorize(long long N) {
    vector<pair<long long, long long> > res;
    for (long long a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        long long ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({a, ex});
    }

    // 最後に残った数について
    if (N != 1) res.push_back({N, 1});
    return res;
}



int main(){
    bool a[101];
    rep(i,101){
        a[i]=false;
    }
    int x,n;
    cin >> x >> n;
    
    rep(i,n){
        int w;
        cin >> w;
        a[w]=true;
    }

    int mi = x;
    int ma = x;


    for(int i=x;i>=0;i--){
        if(!a[i]){
            mi = i;
            break;
        }
    }

    for(int i=x;i<=100;i++){
        if(!a[i]){
            ma = i;
            break;
        }
    }

    if(abs(x-mi)<=abs(x-ma)){
        cout << mi << endl;
    }else{
        cout << ma << endl;
    }
}