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 pb push_back
#define mp make_pair
#define fi first
#define se second
const ll mod = 1e9 + 7;
ll powmod(ll a, ll b) { ll res=1; a%=mod; assert(b>=0); for(;b;b>>=1) { if(b&1) res=res*a%mod; a=a*a%mod; } return res; }
ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a; }

int main() {
	int t;
	// cin >> t;
    t = 1;

	while(t--) {
		int x, n;
        cin >> x >> n;
        
        vector<int> arr(n);
        for(auto &v: arr)
            cin >> v;
        
        if(n == 0) {
            cout << x << endl;
            return 0;
        }
        
        int dif = 0;
        while(true) {
            if(find(arr.begin(), arr.end(), x-dif) == arr.end()) {
                cout << (x-dif) << endl;
                return 0;
            }
            if(find(arr.begin(), arr.end(), x+dif) == arr.end()) {
                cout << (x+dif) << endl;
                return 0;
            }
            
            ++dif;
        }
	}

	return 0;
}