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)

/* https://codeforces.com/contest/1360/problem/H */
#include<bits/stdc++.h>

using namespace std;
using ll = long long;
using vs = vector<string>;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vpi = vector<pair<int, int>>;
using vpl = vector<pair<ll, ll>>;
using ld = double;

#define f first
#define s second 
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define pb push_back
#define	lb lower_bound
#define	ub upper_bound
#define	sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
#define ins insert
#define mp make_pair

const int MAX = 1e5;

int main() {
	int x, n;
	cin>>x>>n;
	vi a(n);
	vi c(n);
	if (n == 0) {
		cout<<x<<endl;
		return 0;
	}

	map<int, bool> cnt;
	for (int i = 0;i < n;++i) {
		cin>>a[i];
		cnt[a[i]] = true;
	}

	vector<pair<int, int>> ans;
	int c1;
	for (int i = 0;i <= 101;++i) {
		if (cnt.find(i) == cnt.end()) {
			c1 = i - x;
			ans.pb(mp(abs(c1), c1));
		}
	}

	sort(all(ans));
	printf("%d\n", ans[0].s + x);
	return 0;
}