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)

#ifdef LOCAL
//#define _GLIBCXX_DEBUG
#endif
//#pragma GCC target("avx512f,avx512dq,avx512cd,avx512bw,avx512vl")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<int, int> Pi;
typedef vector<ll> Vec;
typedef vector<int> Vi;
typedef vector<string> Vs;
typedef vector<P> VP;
typedef vector<vector<ll>> VV;
typedef vector<vector<int>> VVi;
typedef vector<vector<vector<ll>>> VVV;
typedef vector<vector<vector<vector<ll>>>> VVVV;

#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define PER(i, a, b) for(ll i=(a); i>=(b); i--)
#define rep(i, n) REP(i, 0, n)
#define per(i, n) PER(i, n, 0)
const ll INF=1e18+18;
const ll MOD=1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl;
#define ALL(v) v.begin(), v.end()
#define rALL(v) v.rbegin(), v.rend()
#define pb(x) push_back(x)
#define mp(a, b) make_pair(a,b)
#define Each(a,b) for(auto &a :b)
#define rEach(i, mp) for (auto i = mp.rbegin(); i != mp.rend(); ++i)
#ifdef LOCAL
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbgmap(mp) cerr << #mp << ":"<<endl; for (auto i = mp.begin(); i != mp.end(); ++i) { cerr << i->first <<":"<<i->second << endl;}
#define dbgarr(n,m,arr) rep(i,n){rep(j,m){cerr<<arr[i][j]<<" ";}cerr<<endl;}
#define dbgdp(n,arr) rep(i,n){cerr<<arr[i]<<" ";}cerr<<endl;
#define dbgmint(n,arr) rep(i,n){cerr<<arr[i].x<<" ";}cerr<<endl;
#define dbgarrmint(n,m,arr) rep(i,n){rep(j,m){cerr<<arr[i][j].x<<" ";}cerr<<endl;}
#else
#define dbg(...)
#define dbgmap(...)
#define dbgarr(...)
#define dbgdp(...)
#define dbgmint(...)
#define dbgarrmint(...)
#endif
#define out(a) cout<<a<<endl
#define out2(a,b) cout<<a<<" "<<b<<endl
#define vout(v) rep(i,v.size()){cout<<v[i]<<" ";}cout<<endl
#define Uniq(v) v.erase(unique(v.begin(), v.end()), v.end())
#define fi first
#define se second

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }

template<typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; }

template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,v.size())i>>v[j];return i;}

// vector
template<typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
    int len=v.size();
    for(int i=0; i<len; ++i) {
        s<<v[i];
        if(i<len-1) s<<" ";
    }
    return s;
}

// 2 dimentional vector
template<typename T>
ostream &operator<<(ostream &s, const vector<vector<T> > &vv) {
    int len=vv.size();
    for(int i=0; i<len; ++i) {
        s<<vv[i]<<endl;
    }
    return s;
}

//mint
struct mint {
    ll x; // typedef long long ll;
    mint(ll x=0):x((x%MOD+MOD)%MOD){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += MOD-a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= MOD; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime MOD
    mint inv() const { return pow(MOD-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
typedef vector<mint> Vmint;

struct combination {
    vector<mint> fact, ifact;
    combination(int n):fact(n+1),ifact(n+1) {
        assert(n < MOD);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n]*ifact[k]*ifact[n-k];
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<std::setprecision(10);

    ll x,n;
    cin>>x>>n;
    vector<ll> p(n);
    for(ll i=0; i<n; i++) {
    	cin>>p[i];
    }
    ll ans = 0;
    ll sa = 100;
    sort(p.begin(), p.end(), greater<ll>());
    for (ll i = 100; i > 0; i--) {
        auto result = std::find(p.begin(), p.end(), i);
        if (result == p.end()){
            if(sa >= llabs(x-i)) {
                sa = llabs(x-i);
                ans = i;
            }
        }
    }

    cout<<ans<<endl;
    return  0;
}