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 <algorithm>
#include <vector>
using namespace std;

bool have(int x , vector<int> &v) {
  bool ans = false;
  for (int i = 0 ; i < v.size() ; i ++) {
    if (v[i] == x) {
      ans = true;
      break;
    }
  }
  return ans;
}

int main(){
  int x , n , p ;
  cin >> x >> n;
  vector<int>  v(200);
  for (int i = 0 ; i < 200 ; i ++) {
    v[i] = i;
  }
  for (int i = 0 ; i < n ; i ++) {
    cin >> p;
    v.erase(remove(v.begin(), v.end(), p), v.end());
  }

  for (int i = 0 ; i < 300 ; i ++) {
    if (have(x - i , v)) {
      cout << x - i << endl;
      break;
    } else if (have(x + i , v)) {
      cout << x + i << endl;
      break;
    }
  }
}