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>
#define fi first
#define se second
#define pb push_back
using namespace std;
long long n,x,arr[105],kcl;
bool ada[105];
int main(){
	cin>>x>>n;

	for(int i =1;i<=n;i++){
		cin>>arr[i];
		ada[arr[i]]=1;
	}
	kcl=1e9;
	if(!ada[x]){
		kcl=x;
	}
	
	for(int i =1;i<=n;i++){
		if(!ada[arr[i]+1]){
			if(abs(arr[i]+1-x)<=abs(kcl-x)){
				if(abs(arr[i]+1-x)==abs(kcl-x)){
					if(arr[i]+1<kcl)kcl=arr[i]+1;
				}
				else kcl=arr[i]+1;
			}
		}
		if(!ada[arr[i]-1]){
			if(abs(arr[i]-1-x)<=abs(kcl-x)){
				if(abs(arr[i]-1-x)==abs(kcl-x)){
					if(arr[i]-1<kcl)kcl=arr[i]+1;
				}
				kcl=arr[i]-1;
			}
		}
	}
	cout<<kcl<<endl;
}