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;

int main(){
    int X, N, near=0, dif, count=0, ans1, ans2, ans=0;
    cin >> X >> N;

    if(N == 0){
        cout << X << endl;
        return 0;
    }

    vector<int> P(N);

    for(int n=0; n<N; n++){
        cin >> P.at(n);
    }

    for(int i=0; i<N; i++){
        count = 0;
        for(int n=0; n<N && count != 2; n++){
            dif = X - P.at(n);
            if(abs(dif) == near){
                ans1 = dif;
                count++;
            } else if(abs(dif) == near && count == 1) {
                count++;
            }
        }
        if(count == 0){
            ans = X - near;
            break;
        } else if(count == 1 && ans1 != 0){
            ans = X + ans1;
            break;
        }
        near++;
    }
    cout << ans << endl;
    return 0;
}