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>
using namespace std;
#define ll long long
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL)
#define endl "\n"
#define pb push_back
#define mod 1000000007
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
int dx8[]={1,-1,0,0,1,1,-1,-1};
int dy8[]={0,0,1,-1,1,-1,1,-1};
ll binexp(ll a, ll b, ll m) {
	a %= m;
	ll res = 1;
	while (b > 0) {
		if (b & 1)
			res = res * a % m;
		a = a * a % m;
		b >>= 1;
	}
	return res;
}
ll modinvfermat(ll a, ll m)
{
	return binexp(a, m - 2, m);
}
void task(bool flag)
{
	if (flag)
		cout << "YES\n";
	else
		cout << "NO\n";
}
ll lcm(ll a, ll b)
{
	return ((1LL * a * b) / (__gcd(a, b)));
}



//ll a[]={6,8,4,2,6};

//ll tb=100;
void solve(){
	ll x,n;
	cin>>x>>n;
	vector<ll> v(n);
	unordered_map<ll,ll> m;

	for(ll i=0;i<n;i++){
		cin>>v[i];
		m[v[i]]=1;
	}
	bool check=false;
	ll ans=0;
	ll diff=0;
	while(!check){
		if(m[x-diff]==0){
			ans=x-diff;
			break;
		}
		else if(m[x+diff]==0){
			ans=x+diff;
			break;
		}
		else{
			diff++;

		}
	}
	cout<<ans<<endl;
	



		
}



int main()
{  
#ifndef ONLINE_JUDGE
	// for getting input from input.txt
	freopen("input.txt", "r", stdin);
	// for writing output to output.txt
	freopen("output.txt", "w", stdout);
#endif
    fast;
    ll t;
    t=1;
   // cin>>t;
    while(t--){
    	solve();
    }
    
   
  



#ifndef ONLINE_JUDGE
	cout << "\nTime Elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " sec\n";
#endif
return 0;
}