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;

#define pb push_back
#define ll long long int
#define mpi map<int, int>
#define vi vector<int>
#define qi queue<int>
#define si set<int>
#define pii pair<int, int>
#define vpi vector<pii>
#define vs vector<string>
#define pri priority_queue<int>
#define mp make_pair
#define ff first
#define ss second
#define mod 1000000007
#define For(i, n) for (int i = 0; i < n; i++)
#define Ford(i, a, b) for (int i = a; i >= b; i--)
#define Rep(i, x, y) for (int i = x; i <= y; i++)
#define mem(a, b) memset(a, b, sizeof(a))
#define fastio                    \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0)

double tick()
{
    static clock_t oldt, newt = clock();
    double diff = 1.0 * (newt - oldt) / CLOCKS_PER_SEC;
    oldt = newt;
    return diff;
}

ll fastExpo(ll a, ll n)
{
    ll res = 1;
    while (n)
    {
        if (n % 2)
            res = res * a;
        n /= 2;
        a = a * a;
    }
    return res;
}

int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};

int a[110], c[110], mn = 1e9;

int main()
{
    int x, n, res = 0;
    cin >> x >> n;
    mem(c, 0);
    For(i, n)
    {
        cin >> a[i];
        c[a[i]] = 1;
    }

    For(i, 102)
    {
        if (c[i] == 0 && abs(i - x) < mn)
        {
            res = i;
            mn = abs(i - x);
        }
    }

    if (mn == 1e9)
        res = -1;
    cout << res << endl;

    cerr << "execution time is: " << tick() << "\n";
}