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 <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <numeric>
#include <set>
#include <queue>
#include <list>
using namespace std;
typedef long long ll;

void solve() {
	int x, n; cin >> x >> n;
	vector<int> v(n);
	map<int, bool> m;
	for (int &i : v)cin >> i, m[i] = 1;
	vector<int> temp;
	for (int i = 0; i <= 300; i++)if (!m[i])temp.push_back(i);
	int ans = 0, cur = 100;
	sort(temp.rbegin(), temp.rend());
	for (int i : temp) {
		if (abs(i - x) <= cur) {cur = abs(i - x); ans = i;}
	}
	cout << ans;



}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	// int t; cin >> t;
	// while (t--)
	solve();

}