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)

#define TEMPLATE_USED
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pl;
typedef vector<pl> vp;
const ll INF=1001001001;
const ll LINF=1001001001001001001;
const ll D4[]={0,1,0,-1,0};
const ll D8[]={0,1,1,0,-1,-1,1,-1,0};
#define _overload3(_1,_2,_3,name,...) name
#define _rep2(i,n) for(ll i=0;i<(n);++i)
#define _rep3(i,a,b) for(ll i=(ll)(a);i<(ll)(b);++i)
#define rep(...) _overload3(__VA_ARGS__,_rep3,_rep2)(__VA_ARGS__)
#define _repe2(i,n) for(ll i=0;i<=(ll)(n);++i)
#define _repe3(i,a,b) for(ll i=(ll)(a);i<=(ll)(b);++i)
#define repe(...) _overload3(__VA_ARGS__,_repe3,_repe2)(__VA_ARGS__)
#define _rrep2(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define _rrep3(i,a,b) for(ll i=(ll)(b)-1;i>=(ll)(a);i--)
#define rrep(...) _overload3(__VA_ARGS__,_rrep3,_rrep2)(__VA_ARGS__)
#define _rrepe2(i,n) for(ll i=(ll)(n);i>=0;i--)
#define _rrepe3(i,a,b) for(ll i=(ll)(b);i>=(ll)(a);i--)
#define rrepe(...) _overload3(__VA_ARGS__,_rrepe3,_rrepe2)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
template<class T>
bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T>
bool chmin(T &a,const T &b){if(b<a){a=b;return 1;}return 0;}

ll X;
ll N;
vl p;

void solve();

int main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    cin >> X;
    cin >> N;
    p = vl(N);
    for(int i = 0; i < N; ++i){
        cin >> p[i];
    }
    solve();
}

// 1 <= X <= 100
// 0 <= N <= 100
// 1 <= p_i <= 100
// p_1, \ldots, p_N はすべて異なる。
// 入力中のすべての値は整数である。

// ll X, ll N, vl p
// 答えを出力せよ。
void solve(){
    set<ll> st;
    rep(i,N)st.insert(p[i]);
    
    ll ans=-1;
    ll val=INF;
    repe(i,101){
        if(abs(X-i)>=val)continue;
        if(st.count(i))continue;
        ans=i;
        val=abs(X-i);
    }
    cout<<ans<<endl;
}