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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
int checkExt(int k);
vector<int> p;
int n;
int main(){
cin.tie(0);
int x,min,result;
 result=100;

cin >>x >>n;
min=100000;
if(n==0){cout<<x;return 0;}
for(int i=1;i<=n;i++){
 int k;
 cin >>k;
 k=k;
 p.push_back(k);
}
 sort(p.begin(),p.end());

for(int k=-100;k<=100;k++){
 
  if(checkExt(k)==0){
   //printf("k:%d p[c]:%d min:%d result:%d\n",k,p[c],min,result);
  if((k-x)*(k-x)<min){result=k;min=(k-x)*(k-x);}
   else if((k-x)*(k-x)==min&&result>=k){result=k;min=(k-x)*(k-x);}
  }

}
  

cout<<result;

return 0;
}

int checkExt(int k){
 for(int i=0;i<n;i++){
   if(p[i]==k)return 1;
 }
 return 0;
}