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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

long long test_cases;
int main()
{
    
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);

    /* cin >> test_cases; */
    test_cases = 1;

    int x, n;
    set<int> p;


    while (test_cases > 0)
    {
        cin >> x >> n;
        for (int i = 0; i < n; i++)
        {
            int pi;
            cin >> pi;
            p.insert(pi);
        }

        int pos_first = x + 1;
        for (int i = x + 1; p.find(i) != p.end(); i++)
        {
            
            pos_first = i + 1;
        }
        /* cout << "pos_first: " << pos_first << endl; */
        int neg_first = x - 1;
        for (int i = x - 1; p.find(i) != p.end(); i--)
        {
            neg_first = i - 1;
        }
        /* cout << "neg_first: " << neg_first << endl; */
        if (n != 0)
            if ((abs(pos_first - x) == abs(neg_first - x)))
            {
                cout << min(pos_first, neg_first) << endl;
            }
            else
                if (abs(pos_first - x) < abs(neg_first - x))
                    cout << pos_first << endl;
                else
                    cout << neg_first << endl;
        else
            cout << x << endl;
        test_cases--;
    }

    
    return 0;
}