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 <iostream>
#include<bits/stdc++.h>
using namespace std;
int main() {
    int a,s,temp;
    cin >> a >> s;
    vector <int> b;
    for(int i=0;i<s;i++){
        cin >> temp;
        b.push_back(temp-a);
    }
    sort(b.begin(),b.end());
    //
    temp = 0;
    while(temp < s && b[temp]!=0) temp++;

    if (temp >=s) {cout << a;return 0;}

    int i=0,j=0;
    
    while(temp-i>=0 && temp+j<s){
        
        if(b[temp-i]!=-1*i) {
            cout << a-i;
            return 0;
        }
        else if(b[temp+j]!=j) {
            cout << a+j;
            return 0;
        }
        i++;j++;
        
    }
    if(temp-i>=0 && b[temp-i]!=-1*i) cout << a-i;
    else if(temp-i<0) cout << a-i;
    else if(temp+j>=s) cout << a+j;
    else cout << a+j;
    return 0;
}