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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#include <complex>
#include <stdio.h>
#include <time.h>
#include <numeric>
#include <unordered_map>
#include <unordered_set>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
# define ll int64_t
# define str string
# define rep(i,n) for(ll i=0;i<n;i++)
# define rrep(i,n) for(ll i=1;i<=n;i++)
# define ALL(x) (x).begin(), (x).end()
# define SZ(x) ((int)(x).size())
# define pb push_back
# define mod 1000000007
# define PI 3.141592653589793
# define vec vector
#define dump(x) cerr<<#x<<"="<<x<<endl
using namespace std;



#define INF 2000000000
#define MAX_V 10



bool compare_by_b(pair<string,ll> a,pair<string,ll> b){
    if(a.second != b.second) return a.second<b.second;
    else return a.first<b.first;
}

bool my_compare(pair<ll,ll> a,pair<ll,ll> b){
    if(a.first != b.first) return a.first<b.first;
    if(a.second != b.second) return a.second>b.second;
    else return true;
}

ll modpow(ll a,ll n,ll mod1) {
    ll res=1;
    while(n>0){
        if(n&1) res=res*a%mod1;
        a = a*a%mod1;
        n >>= 1;
    }
    return res;
}

ll factorial(ll n){
    ll x=1;
    rrep(i,n) x*=i;
    return x;
}

ll gcd(ll a,ll b) { return b ? gcd(b, a%b) : a; }

ll func(ll a,ll b,ll& ans_x,ll& ans_y){
    if(b == 0){
        ans_x = 1;
        ans_y = 0;
        return a;
    }else{
        ll ret = func(b,(a%b),ans_y,ans_x);
        ans_y -= (a/b)*ans_x;
        return ret;
    }
}

ll lcm(ll a,ll b){
   return a/gcd(a,b)*b;
}
const ll MAX = 5100000;
const ll MOD = 1000000007;
 
ll fac[MAX], finv[MAX], inv[MAX];
 
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
ll COM(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

using graph = vector<vector<ll>>;
pair<pair<ll,ll>,ll> p[100010];

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ll x,n;
    cin>>x>>n;
    if(n==0){
        cout<<x<<endl;
        return 0;
    }
    if(n==100){
        cout<<101<<endl;
        return 0;
    }
    vector<ll> p(101,0),a(n);
    rrep(i,100) p[i]=i;
    rep(i,n){
        cin>>a[i];
        p[a[i]]=0;
    }
    ll mini=x+1,ans=x+1;
    rrep(i,100){
        if(p[i]!=i) continue;
        else{
            if(mini>abs(x-p[i])){
                ans=i;
                mini=abs(x-p[i]);
        }
      }
    }
    cout<<ans<<endl;
    return 0;
}