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>
using namespace std;

typedef long long int LLI;

void solve()
{
    int x,n;
    cin>>n>>x;
    if(!x){
        cout<<n;
        return;
    }
    int p[x];
    memset(p,1,sizeof(p));
    for(int i=0;i<x;++i){
        int num;
        cin>>num;
        p[num]=0;
    }
    int mn=10000007,ans;
    for(int num=-100;num<=100;++num){
        if(p[num]){
            if(abs(n-num)<mn){
                mn=abs(n-num);
                ans=num;
            }
        }
    }
    cout<<ans;
    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
    return 0;
}