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>                                           
 
#define int long long
//#pragma GCC optimize("Ofast")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
 
 
#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);
#define forev(i, b, a) for(int i = (b); i >= (a); --i)
#define forn(i, a, b) for(int i = (a); i <= (b); ++i)
#define all(x) x.begin(), x.end()
#define sz(s) (int)s.size()
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define s second
#define f first
 
 
using namespace std;



typedef pair < long long, long long > pll;
typedef unsigned long long ull;
typedef pair < int, int > pii;
typedef vector < pii > vpii;
typedef vector < int > vi;
typedef long double ldb;
typedef long long ll;
typedef double db;
 
 
const int inf = 1e9, maxn = 4e5 + 148, mod = 1e9 + 7, N = 511;
const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 466;
const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);
const db eps = 1e-12, pi = 3.14159265359;
const ll INF = 1e18;


int x, n, p[N];
bool was[N];

main () {
	cin >> x >> n;
	for (int i = 1; i <= n; ++i) {
		cin >> p[i];
		was[p[i]] = 1;
	}

	pii ans = {inf, inf};
	for (int y = -N; y < N; ++y)
		if (y < 0 || !was[y])
			ans = min(ans, mp(abs(x - y), y));
	cout << ans.s << endl;
}