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 <algorithm>
#include <iostream>
#include <vector>
#include <numeric>
#include <string>
#include <stdio.h>
#include <set>
#include <iomanip>
using namespace std;
 
 
int main(){
    int X;
    int N;
    int tmp;
    int ans = 0;
    cin >> X >> N;
    if (N == 0){
        cout << X << endl;
        return 0;
    }
    vector<int> p(N);
    vector<int> a(101, -1);
    for (int i=0; i<101; i++){
        a[i] = i;
    }
    for (int i=0; i<N; i++){
        cin >> tmp;
        a[tmp] = -100;
    }
    for (int i=0; i<101; i++){
        if (X-ans > abs(X-a[i])) ans = a[i];
    }
    cout << ans << endl;

    return 0;
}