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 <iostream>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <string>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <cstdlib>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vp = vector<pii>;
using vs = vector<string>;
using qi = queue<int>;
using qp = queue<pii>;
using pqi = priority_queue<int>;
using pqim = priority_queue<pii, vector<pii>, greater<pii> >;
using ppp = pair<int, pair<int, int>>;
#define all(vec) vec.begin(), vec.end()
#define rall(vec) vec.rbegin(), vec.rend()
#define pb push_back
const ll mod = 998244353;
const ll inf = 1e18;

int x, n;
vi a;

void input() {
	cin >> x >> n;
	a = vi(n);
	for (int i = 0; i < n; ++i)
		cin >> a[i];
}

void solve() {

	if (!n) {
		cout << x;
		return;
	}

	int ex[110] = {};
	for (int i = 0; i < n; ++i) {
		ex[a[i]] = 1;
	}

	int l = x, r = x;
	while (ex[l]) l--;
	while (ex[r]) r++;

	if (x - l <= r - x) cout << l;
	else cout << r;



	return;
}

int main() {
	cin.tie(NULL); ios_base::sync_with_stdio(false);

	input();
	solve();
}