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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
int main(){
int x,n,o=0,k,p;
cin>>x>>n;
int z=x+1,l=x-1;
int y[n];
if (n==0)
    cout<<x;
else{
for(int i=0;i<n;i++)
cin>>y[i];
sort(y,y+n);
for(int i=0;i<n;i++){
if(x!=y[i])
o++;
else{
k=p=i;
break;}
}
int u;
if(n-k>=k)
    u=k;
else
    u=n-k;
if (o==n)
cout<<x;
else
for(int i=0;i<u;i++){

if(y[k]!=l){
cout<<l;
i++;
break;
}
else if(y[p]!=z){
cout<<z;
i++;
break;}
else{
z++;
l--;
k--;
p++;
}
}
}
return 0;
}