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>
#define rep(i,n) for (int i=0; i< int (n); i++)
using namespace std;
typedef long long ll;

int main(){

int x, n;
cin >> x >> n;
vector<int> p(110);
rep(i,n) cin >> p[i+1];

if(n==0){
    cout << x << endl;
    return 0;
}

for(int i=0; i<=110; i++){
    bool aru=false;
    for(int j=1; j<=n; j++) {
        if(x-i == p[j]) {
        aru=true;
        }
    }
    if(!aru) {
        cout << x-i << endl;
        return 0;
    }
    aru=false;
    for(int j=1; j<=n; j++) {
        if(x+i == p[j]) {
        aru=true;
        }
    }
    if(!aru) {
        cout << x+i << endl;
        return 0;
    }
}

}