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>

using namespace std;

int main()
{
    int x, n;

    vector<int> p;

    cin >> x >> n;

    for (int i = 0; i < n; ++i)
    {
        p.emplace_back();
        cin >> p.back();
    }

    for (int i = 0;; ++i)
    {
        if(x - i < 0)
        {
            cout << 0 << "\n";
            return 0;
        }

        auto itr = find(p.begin(), p.end(), x - i);
        if (itr == p.end())
        {
            cout << x - i << "\n";
            return 0;
        }
        
        itr = find(p.begin(), p.end(), i + x);
        if (itr == p.end())
        {
            cout << i + x << "\n";
            return 0;
        }

    }

        return 0;
}