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 x,n,a[101],mi=0x3f3f3f3f,pos;
bool find(int x)
{
	for(int i=1;i<=n;i++)
	{
		if(x==a[i])return false;
	}
	return true;
}
int main()
{
	cin>>x>>n;
	int f=1;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		if(a[i]==x)f=0;
	}
	if(f)cout<<x,exit(0);
	for(int i=1;i>=1;i++)
	{
		if(find(i))
		{
			if(abs(x-i)<mi)
			{
				mi=abs(x-i);
				pos=i;
			}
			else break;
		}
	}
	cout<<pos;
	
}