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)

// Online C++ compiler to run C++ online.
// Write C++ code in this online editor and run it.

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main() {
    
    int x,n;
    cin>>x>>n;
    int ans1,ans2;
    vector<int> p(n,0);
    for(int i=0;i<n;i++){
        cin>>p[i];
    }
    sort(p.begin(), p.end());
    ans1 = x-1;
    for(int i = n-1; i >= 0; i--){
        if(p[i] == ans1){
            ans1 = ans1-1;
        }
    }
    ans2 = x+1;
    for(int i = 0; i < n; i++){
        if(p[i] == ans2){
            ans2 = ans2+1;
        }
    }
    if(n==0){
        cout<<x;
    }
    else if(abs(ans1-x) < abs(ans2-x)){
        cout<<ans1;
    }
    else if(abs(ans1-x) == abs(ans2-x)){
        cout<<min(ans1,ans2);
    }
    else{
        cout<<ans2;
    }
    return 0;
}