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: Runtime Error)

#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <string>
#include <cmath>
#include <iostream>
#include <stack>
#include <queue>
#include <climits>
#include <iomanip>

#define all(x) x.begin(), x.end()
#define MOD 998244353

const long double PI = 3.141592653589793236L;

typedef long long int ll;
typedef long double ld;

using namespace std;

void solve()
{
	// Your code goes here.
	int n, r;
	cin>>r>>n;
	vector<int>v(n);
	set<int>s;

	for (int i=1; i<=n; i++)
    {
        cin>>v[i];
        s.insert(v[i]);
    }
    if (n==0)
    {
        cout<<r<<endl;
        return;
    }
    int i=1, check, nextt, g;
    check = r;
    while (i<=100)
    {
        nextt=(pow((-1), i));
        g=nextt*i;
        check=(check+g);

        if (s.find(check)==s.end())
        {
            cout<<check<<endl;
            return;
        }else
        {
            //cout<<i<<endl;
            //cout<<check<<endl;
            i++;

        }
    }
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int T = 1;
    //cin >> T; // If only one test case then you can comment this line out.

    while (T--)
    {
        solve();
    }

    return 0;
}