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;
 int x1=x;
 vector<int> vec(n);
for(int i=0;i<n;i++){
  cin>>vec.at(i);
}
sort(vec.begin(), vec.end());
for(int i=0;i<n;i++){
  if(vec.at(i)==x){
    x=x+1;
  }
}
int x2=x;
x=x1;
reverse(vec.begin(), vec.end());
for(int i=0;i<n;i++){
  if(vec.at(i)==x){
    x=x-1;
  }
}
if(x2-x1<x1-x){
  x=x2;
}

cout<<x<<endl;
 }