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)

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5+5;
const int MaxN = 2e5+5;
typedef long long ll;
typedef long double ld;

int a[N];
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int x, n;
    cin >> x >> n;
    map<int,int>f;
    for (int i = 0; i < n; i++){
        int b;
        cin >> b;
        f[b]++;
    }
    int l= x, r=x;
    while (true){
        if (f[r] == 0){
            cout << r;
            return 0;
        }else r++;
        if (f[l] == 0){
            cout << l;
            return 0;
        }else l--;
    }
}