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;
using ll = long long;
using Graph = vector<vector<int>>;
using pint = pair<int, int>;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

int main() {
    int x, n;
    cin >> x >> n;
    if (n == 0) {
        cout << x << endl;
        return 0;
    }
    vector<int> p(n);
    for (int i = 0; i < n; ++i) cin >> p[i];
    sort(p.begin(), p.end());
    int res = 1000;
    int res_abs = 1000;
    for (int i = -1; i <= 101; ++i) {
        if (find(p.begin(), p.end(), i) != p.end()) continue;
        int tmp = abs(x - i);
        if (tmp < res_abs) {
            res = i;
            res_abs = tmp;
        }
    }
    cout << res << endl;

    return 0;
}