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 _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<typename T> bool chmax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<typename T> bool chmin(T &a,T b){if(a>b){a=b;return true;}return false;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef tuple<ll,ll,ll> tlll;
typedef tuple<int,int,int> tiii;
const ll mod=1e9+7;
const int inf=1<<30;

int main(){
  int x,n; cin >> x >> n;
  vector<int> p(n);
  rep(i,n) cin >> p[i];
  set<int> st;
  for(int i=-100;i<101;i++){
     st.insert(i+1);
  }
  rep(i,n){
    if(st.count(p[i])) st.erase(p[i]);
  }
  int mnsa=inf;
  int ans=x;
  for(auto v:st){
    if(abs(v-x)<mnsa){
      mnsa=abs(v-x);
      ans=v;
    }
  }
  if(st.count(x)) ans=x;
  cout << ans << endl;
}