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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main(void){
    int x,n;
    cin >> x >> n;
    vector<int> p(n);
    for(int i=0;i<n;i++){
        cin >> p.at(i);
    }
    sort(p.begin(),p.end());
    vector<int> forbid;
    for(int i=0;i<n-1;i++){
        int y=p.at(i+1)-p.at(i);
        if(y!=1){
            for(int j=1;j<y;j++){
                forbid.push_back(p.at(i)+j);
            }
        }
    }
    int f=forbid.size();
    int hint=100;
    int ans=0;
    for(int i=0;i<f;i++){
        int a=abs(x-forbid.at(i));
        if(hint>a){
            hint=a;
            ans=forbid.at(i);
        }
    }
    if(n==0||x>p.at(n-1)||x<p.at(0)){
        cout << x << endl;
    }else{
        cout << ans << endl;
    }
}