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;
#define mod 1000000007
#define pb push_back
#define pob pop_back
#define mp make_pair
#define ff first
#define ss second
#define pi 3.14159265
#define fast ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair <ll, ll> pll;
const ll M=1e9+7;
ll inf=1e18;
int main()
{
    int n,x;
    cin>>x>>n;
    int a[n];
    int d[102]={0};
    for(int i=0;i<n;i++){
        cin>>a[i];
        d[a[i]]=1;
    }
    int ab=INT_MAX,ans=0;
    for(int i=1;i<=101;i++){
        if(d[i]==0){
            if(abs(i-x)<ab){
                ab=abs(i-x);
                ans=i;
            }
        }
    }
    cout<<ans;
}