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)

                  /**Bismillahir Rahmanir Rahim.**/
#include<bits/stdc++.h>
#include<stdlib.h>
#include<iostream>
#include<vector>
#include<math.h>
#include<cstring>
using namespace std;
typedef long long int            ll;
typedef long double              ld;
#define Faster  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pi                 acos(-1);
#define test        int t; cin>>t; while(t--)
//scanf("%[^\n]%*c", str);
//sort(s,s+n,greater<int>());
//cin.getline(array,size);
int main()
{
    int x,n,i,j,a,c,ans1,ans2,p1=0,p2=0;
    cin>>x>>n;
    int b[101]={0};
    if(n==0)
        cout<<x<<endl;
    else
    {
        b[x]=x;
        for(i=1;i<=n;i++)
        {
            cin>>a;
            b[a]=a;
        }
        for(i=x+1;i<=101;i++)
        {
            p1++;
            if(b[i]==0)
            {
                ans1=i;
                break;
            }
        }
        for(i=x-1;i>=0;i--)
        {
            p2++;
            if(b[i]==0)
            {
                ans2=i;
                break;
            }
        }
        if(p1>p2)
        {
            cout<<ans2<<endl;
        }
        else if(p1<p2)
        {
            cout<<ans1<<endl;
        }
        else
        {
            cout<<ans2<<endl;
        }
    }
    return 0;
}