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 fastio ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
#define ll long long
const int N = 2e5 + 1;
//ll int freq[100001];
ll int bigPower(ll int x, unsigned ll int y, ll int p)
{
    ll int res = 1;     // Initialize result

    x = x % p; // Update x if it is more than or
    // equal to p

    if (x == 0) return 0; // In case x is divisible by p;

    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;

        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
void findAns()
{
    int x, n;
    cin >> x >> n;
    int a;
    unordered_set<int> data;

    for (int i = 0; i < n; i++)
    {
        cin >> a;
        data.insert(a);
    }
    int minAns = 2000;
    int ans;
    for (int i = 0; i < 110; i++)
    {
        if (data.find(i) == data.end())
        {
            if (abs(x - i) < minAns)
                ans = i, minAns = abs(x - i);
        }
    }

    cout << ans;


}
int main()
{
// #ifndef ONLINE_JUDGE
//     freopen("in.txt", "r", stdin);
//     freopen("o1.txt", "w", stdout);
// #endif

    fastio;

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

    // while (t--)
    findAns();
}