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)

#include<bits/stdc++.h>
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")

using namespace std;

typedef long long ll;

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>pb_ds;

#define IN freopen("in.txt","r",stdin)
#define OUT freopen("out.txt","w",stdout)
#define pi pair<int,int>
#define F first
#define S second
#define pb push_back
#define pp pop_back
#define eps 10E-10
#define mod 998244353
#define pai acos(-1)
#define N 1005


int main ()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n,x;
    map<int,bool>mp;
    cin>>x>>n;
    for(int i=1;i<=n;i++){
      int ar;cin>>ar;
      mp[ar]=1;
    }
    int mn=INT_MAX,ans;
    for(int i=-100;i<=200;i++){
      if(!mp[i]&&abs(i-x)<mn){
         ans=i;
         mn=abs(i-x);
      }
    }
    cout<<ans<<endl;
    return 0;
}