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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <cmath>
#define rep(i,n) for (int i = 0; i<(n); i++)
using namespace std;
using P = pair<int,int>;
#include <vector>
#include <algorithm>


int main(){
    int x,n;
    cin >> x>>n;
    vector<int> d(102);

    rep(i,n){
        int p;
        cin >> p;
        d[p]=1;
    }
int minNum = 100000;
    for(int i = 0; i<101;i++){
        if (d[i]==1){
            continue;
        }else{
//            cout<<abs(i-x)<<" ";
            minNum = min(abs(i-x), minNum);
        }
    }

    if (d[x - minNum]==1){
        cout << x + minNum;
    }else{
        cout << x - minNum;

    }
//    cout <<endl;
//    for (int i = 0; i < 101; i++) {
//        cout<< d[i]<<" ";
//    }

}