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: Accepted)

#include<bits/stdc++.h>

using namespace std;

#define FAST() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

int main()
{
    FAST();
    bool mark[105]={0};
    int i,n,x,y,z;
    cin >> x >> n;
    while(n--)
    {
        cin >> y;
        mark[y]=1;
    }
    for(i=x-1;i>-1;i--)
    {
        if(!mark[i])
        {
            y=i;
            break;
        }
    }
    for(i=x;i<102;i++)
    {
        if(!mark[i])
        {
            z=i;
            break;
        }
    }
    if(z-x < x-y)cout << z;
    else cout << y;
    return 0;
}