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>

#define F first
#define S second
#define MP make_pair
#define pb push_back
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define LCM(a, b) (a) / __gcd((a), (b)) * (b)
#define CEIL(a, b) (a)/(b)+(((a)%(b))?1:0)
#define ln '\n'

using namespace std;
using LL = long long;
using ldouble = long double;
using P = pair<int, int>;
using LP = pair<LL, LL>;

static const int INF = INT_MAX;
static const LL LINF = LLONG_MAX;
static const int MIN = INT_MIN;
static const LL LMIN = LLONG_MIN;
static const int MOD = 1e9 + 7;
static const int SIZE = 200005;

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

vector<LL> Div(LL n) {
    vector<LL> ret;
    for(LL i = 1; i * i <= n; ++i) {
        if(n % i == 0) {
            ret.pb(i);
            if(i * i != n) ret.pb(n / i);
        }
    }
    sort(all(ret));
    return ret;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int X, N;
    cin >> X >> N;
    vector<int> v(N);
    int t[SIZE] = {};
    for(int i = 0; i < N; ++i) {
        cin >> v[i];
        t[v[i]]++;
    }

    int mn = X + 1;
    int m = -1;
    for(int i = 0; i <= 101; ++i) {
        if(t[i] != 0) continue;
        if(mn > abs(X - i)) {
            mn = abs(X - i);
            m = i;
        }
    }
    cout << m << endl;

    return 0;
}