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: Runtime Error)

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

void test_case()
{
int n,x;
cin>>x>>n;
vector<int> a,b;
int tem=0;
for(int i=1;i<=n;++i)
{
          cin>>tem;
          a.push_back(tem);
}
for(int i=1;i<=n;++i)
{
b.push_back(abs(a[i]-x));
}
int h=*min_element(b.begin(),b.end());
for(int i=1;i<=n;++i)
{
if(abs(a[i]-x)==h)
{
    cout<<a[i]+1<<"\n";
    break;
}
}


}
int main()
{
//ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
        test_case();

    return 0;
    }