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 int long long
#define endl "\n"
#define pb push_back
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define IN freopen("input.txt","r",stdin);
#define OUT freopen("output.txt","w",stdout);

const int N = 2e5 + 5;

void solve();
int32_t main()
{
    IOS;
    int x,n;
    cin >> x >> n;
    set<int>a;
    for(int i=0;i<=101;i++)
        a.insert(i);
    for(int i=0;i<n;i++){
        int t;
        cin >> t;
        a.erase(t);
    }
    auto it = a.lower_bound(x);
    int p = abs(*it-x);
    int q = INT_MAX;
    if(it!=a.begin())
        q = abs(*(--it) - x);
    if(p<q)
        cout << x+p << endl;
    else
        cout << x-q << endl;
}