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)

#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define ll long long
#define endl  '\n'
using namespace std;
const int mod=1e9+7;
const int MAX=1e6+7;
int a[105];
int main()
{
    ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);
  	int x,n,xx;cin>>x>>n;
  	for(int i=0;i<n;i++)
  	{
  		cin>>xx;a[xx]++;	
	}
	if(n==0)cout<<x;
	else{
		int i=x-1,ans,l=0,r=0;
		for(;i>=0;i--)
		{
			if(a[i]==0){
				l=i;break;
			}
		}
		i=x+1;
		for(;i<=100;i++)
		{
			if(a[i]==0){
				r=i;break;
			}
		}
		if(abs(x-r)<abs(x-l)){
			cout<<r;
		}
		else if(abs(x-r)>abs(x-l))
		{
			cout<<l;
		}
		else cout<<l;
	}
	
    return 0 ;
}