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 rep(i, n) for(int i = 0;i < (n);i++)
using namespace std;

int main(){
    int x, n;
    vector<int> p(100);
    cin >> x >> n;
    rep(i, n){
        cin >> p.at(i);
    }
    int j = 0;
    while(n+j < 100){
        p.at(n+j) = 101;
        j++;
    }

    sort(p.begin(), p.end());
    int l = 0;
    int min = 1000;
    rep(i, 102){
        if(i != p.at(l) && abs(x - min) > abs(x-i)){
            min = i;
        } else l++;
        
    }
    cout << min << endl;
    return 0;
}