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;
int main(){
  int x,n;
  cin>>x>>n;
  vector<int> v;
  if(n==0)
    cout<<x;
  else{
  for(int i=0;i<n;i++){
   int c;
    cin>>c;
    v.push_back(c);
  }
    vector<int>::iterator it;
       vector<int>::iterator itr;
   
    int i=-1;
    while(1){
      i++;
      it = std::find (v.begin(), v.end(),x-i); 
       itr = std::find (v.begin(), v.end(),x+i); 
      if (it == v.end()&&itr == v.end()){ 
        cout<<min(x-i,x+i);
        break;
      }
      else if (it == v.end()&&itr != v.end()) {
        cout<<x-i;
        break;
      }
      else if(it!= v.end()&&itr== v.end()){
       cout<<x+i;
        break;
      }
      else
        continue;
   
    }
 
  }
  
  return 0;
}