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 <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
        int x, n, ans = -1;
        cin >> x >> n;
        vector<int> arr(n);
        for(int i = 0 ; i<n ; i++) {
                cin >> arr[i];
        }
        ans = x;
        if(n > 0){
                int mx = *max_element(arr.begin(), arr.end());
                int mn = *min_element(arr.begin(), arr.end());
                vector<int> p;
                for(int i = mn ; i<=mx ; i++) {
                        if(find(arr.begin(), arr.end(), i) == arr.end()) {
                                p.push_back(i);
                        }
                }
                int d = abs(x - p[0]);
                ans = p[0];
                for(int i = 1 ; i<p.size() ; i++){
                        if(d > abs(x-p[i])){
                                d = abs(x-p[i]);
                                ans = p[i];
                        }
                }
        }
        cout << ans << "\n";
        return 0;
}