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;

// const double PI = acos(-1.0); 

void solve();

void fastio()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout); 
    #endif
    
    fastio(); 

    // int t; 
    // cin >> t;

    // while (t--) 
    solve(); 

    return 0;
}

void solve()
{
    int X, N;
    cin >> X >> N;
    set <int> sequence;
    for (int i = 0; i < N; ++i)
    {
        int input;
        cin >> input;
        sequence.insert(input);
    }

    int i = 0, ans;
    bool ansFound = false;
    while (not ansFound)
    {
        if (sequence.find(X - i) == sequence.end())
        {
            ansFound = true;
            ans = X - i;
            break;
        }
        else if (sequence.find(X + i) == sequence.end())
        {
            ansFound = true;
            ans = X + i;
            break;
        }
        i++;
    }

    cout << ans << "\n";
}