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: Accepted)

#include<iostream>
#include<algorithm>
#include<unordered_map>
using namespace std;
int main(){
int x,n;
cin>>x>>n;
unordered_map<int,int>map;
int* arr=new int[n];
for(int i=0;i<n;i++){
    cin>>arr[i];
    map[arr[i]]=1;
}
int k=1;
while(1){
    if(map.count(x)==0){
        cout<<x<<endl;
        break;
    }
    if(x-k>=0)
   if(map.count(x-k)==0){
       cout<<x-k<<endl;
       break;
   }
    if(map.count(x+k)==0){
       cout<<x+k<<endl;
       break;
   }
   k++;
  }
 
}