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;

typedef int64_t Int;
#define all(x) (x).begin(), (x).end()
 
const double EPS = 1e-10;
const Int INF = 1e18;
const int inf = 1e9;
const Int mod = 1e9+7;

bool print_space_enable = false;
void print() { 
    cout << endl; 
    print_space_enable = false;
}

template <class Head, class... Tail>
void print(Head&& head, Tail&&... tail) {
    if (print_space_enable) cout << " ";
    cout << fixed << setprecision(15) << head;
    print_space_enable = true;
    print(std::forward<Tail>(tail)...);
}

template<typename T>
void print(vector<T> v) {
    for (size_t i = 0; i < v.size(); i++) {
        if (i > 0) std::cout << " ";
        std::cout << v[i];
    }
    std::cout << std::endl;
}

int main() {
    Int x, n;
    cin >> x >> n;
    vector<Int> p(n);
    vector<bool> a(101, false);
    for (Int i = 0; i < (Int)n; i++) {
        cin >> p[i];
        a[p[i]] = true;
    }
    if (not a[x]) {
        print(x);
        return 0;
    }
    Int i = 1;
    while (true) {
        Int res = INF;
        if (x - i < 1) {
            res = min(res, x - i);
        }
        if (x + i > 100) {
            res = min(res, x + i);
        }
        if (not a[x - i]) {
            res = min(res, x - i);
        }
        if (not a[x + i]) {
            res = min(res, x + i);
        }
        if (res != INF) {
            print(res);
            return 0;
        }
        i++;
    }
    return 0;
}