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 mod 1000000007
#define pb push_back
#define ppb pop_back
#define cs(n) cout<<n<<" "
#define cn(n) cout<<n<<"\n"
#define rep(i,j,k) for(int i=j;i<k;i++)
#define rrep(i,j,k) for(int i=j;i>=k;i--)
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
ll bin(ll a, ll b)
{
	if (b == 0)
		return 1;
	if (b % 2 == 0)
		return bin((a * a) % mod , b / 2) % mod;
	return ((a % mod) * (bin((a * a) % mod , b / 2) % mod)) % mod;
}
void solve()
{
	int x, n;
	cin >> x >> n;
	map<int , int>mp;
	vector<int>v;
	rep(i, 0, n) {
		int a;
		cin >> a;
		mp[a]++;
	}
	rep(i, -2, 103) {
		if (mp[i] == 0) {
			v.pb(i);
		}
	}
	sort(v.begin(), v.end());
	auto it = lower_bound(v.begin(), v.end(), x);
	if (it == v.end()) {
		it--;
		cout << (*it);
		return;
	}
	auto it1 = it;
	it1--;
	if (x - (*it1) <= (*it) - x) {
		cout << (*it1);
	}
	else {
		cout << (*it);
	}
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r" , stdin);
	freopen("output.txt" , "w" , stdout);
#endif
	fastio;
	int t;
	//cin>>t;
	t = 1;
	while (t--)
	{
		solve();
	}
	return 0;
}