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 ll long long
#define ld long double
#define pb(a) push_back(a)
#define pbp(a,b) push_back({a,b})
#define rep(p,a,b)    for(int p=a;p<b;p++)
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
#define INF 0x3f3f3f3f

ll nextPowerOf2(ll n)   
{ 
    n--; 
    n |= n >> 1ll; 
    n |= n >> 2ll; 
    n |= n >> 4ll; 
    n |= n >> 8ll; 
    n |= n >> 16ll; 
    n++; 
    return n; 
}

bool cmp(ll a,ll b)
{
    return (a>b);
}

int main() {
    
    #ifndef ONLINE_JUDGE
    // for getting input from input.txt
    freopen("input.txt", "r", stdin);
    // for writing output to output.txt
    freopen("output.txt", "w", stdout);
    #endif

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    ll n,i,j,x,y;

    cin>>x>>n;
    map<int,int>m;

    rep(p,0,n)
    {
        cin>>y;
        m[y]=1;
    }

    i=j=x;
    while(m[i] && m[j]){i--;j++;}

    if(!m[i])cout<<i;
    else cout<<j;
    return 0;
}