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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <map>
#include <iomanip>
#include <stdio.h>
#include <bitset>

typedef long long LL;

constexpr long long INFLL = 1e18;

using namespace std;

int main()
{
    int x, n;
    cin >> x >> n;
    vector<int> p(n);
    for (int i = 0; i < n; i++)
    {
        cin >> p.at(i);
    }
    for (int i = 0; i <= n; i++)
    {
        if (find(p.begin(), p.end(), x - i) == p.end())
        {
            cout << x - i << endl;
            return 0;
        }

        if (find(p.begin(), p.end(), x + i) == p.end())
        {
            cout << x + i << endl;
            return 0;
        }
    }

    return 0;
}