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;
#define ll long long int 
#define lp(x,y) for(ll i=x;i<y;i++)
#define pb push_back
 
 
  
// Driver code 
int main() 
{ 
    int x,n;
    cin>>x>>n;
    int arr[n],ar[101]={0};
    if(n!=0)
    {
    lp(0,n) {cin>>arr[i]; ar[arr[i]]++;}
    }
    
    sort(arr,arr+n);
    if(ar[x]==0) cout<<x;
    else
    {
        int j=1;
        for(int i=x;i<=100;i++)
        {
            if(ar[x-j]==0) {cout<<x-j; break;}
            else if(ar[x+j]==0) { cout<<x+j; break;}
            j++;
        }
    }
    
}