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>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std ;
#define int                 long long
#define ull                 unsigned long long int
#define ld                  long double
#define vi                  vector<int>
#define vii                 vector<pair<int,int>>
#define vvi                 vector<vector<int>>
#define pii                 pair<int,int>
#define prec(x)             fixed<<setprecision(15)<<x
#define all(x)              x.begin(),x.end()
#define rall(x)             x.rbegin(),x.rend()
#define rev(x)              reverse(x.begin(),x.end())
#define mset(a,b)           memset(a,b,sizeof(a))
#define mcpy(a,b)           memcpy(a,b,sizeof(a))
#define fo0(i,n)            for(int i=0,i##end=n;i<i##end;i++)
#define fo1(i,n)            for(int i=1,i##end=n;i<=i##end;i++)
#define fo(i,a,b)           for(int i=a,i##end=b;i<=i##end;i++)
#define foe(i,x)            for(__typeof((x).end())i=(x).begin();i!=(x).end();++i)
#define fre(i,x)            for(__typeof((x).rend())i=(x).rbegin();i!=(x).rend();++i)
#define mk(arr,n,type)  	type *arr=new type[n];
#define dbg(x) cerr << #x << " is " << x << endl;
#define FAST_IO ios_base::sync_with_stdio(false); cin.tie(0);cout.tie(0)
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }
clock_t beg = clock();
const int inf=1e18;const int mod=1e9+7;const int N=2e5+9;
const double pi = 3.1415926535897932384626433;
void FILE_IO(){
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}

void JH(){
    int d=0,e=0,f=0,g=0,p=0,w=0,flag=0,ans=INT_MAX,res=0,cnt=0,sum=0;double f1=0.0,f2=0.0,f3=0.0;
    int x,n;cin>>x>>n;
    int a[n];
    set<int> s;
    fo0(i,n){
    	cin>>a[i];
    	s.insert(a[i]);
    }
    if(s.size()==0){
    	cout<<x;
    	return;
    }
    for(int i=x+102;i>=x-102;i--){
    	if(s.find(i)==s.end()){
    		if(abs(x-i)<=ans){
    			ans=abs(x-i);
    			res=i;
    		}
    	}
    }
    cout<<res;
}

int32_t main()
{
    FILE_IO();FAST_IO;int t=1;
    // cin>>t;
    while(t--){JH();}cerr<<(double)(clock()-beg)/1000<<"s ";
}