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 <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <tuple>
#include <vector>

using namespace std;
typedef long long ll;

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define reps(i, f, n) for (ll i = (f); i < (n); i++)
#define repr(i, n) for (ll i = n; i >= 0; i--)
#define repv(v) for (auto it = (v).begin(); it != (v).end(); it++)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define pb push_back
#define INIT    \
    cin.tie(0); \
    ios::sync_with_stdio(false);

template <class T>
inline bool chmax(T& a, T b) {
    return a = (a < b) ? b : a;
}
template <class T>
inline bool chmin(T& a, T b) {
    return a = (a > b) ? b : a;
}

ll const INF = 1LL << 60;
ll const MOD = 1000000007;

int main() {
    INIT;

    ll X, N;
    cin >> X >> N;

    vector<ll> p(N);
    rep(i, N) cin >> p[i];

    vector<ll> cnt(105, 0);
    rep(i, N) { cnt[p[i]]++; }

    ll ans = INF;
    ll idx = -1;
    reps(i, 0, 102) {
        if (cnt[i] != 0) continue;

        if (ans > abs(X - i)) {
            ans = abs(X - i);
            idx = i;
        }
    }

    cout << idx << endl;

    return 0;
}