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>
using namespace std;

int main() {
  int X,N;
  cin >> X >> N;
  
  if (N == 0){
    cout << X << endl;
    return 0;
  }
  
  vector<int>P(N);
  for (int i = 0; i < N; i++) {
    cin >> P.at(i);
  }
  
  vector<int>p(102,200);
  for (int i = 0; i < 102; i++) {
      for (int j = 0; j < N; j++) {
        if (P.at(j) == i){
          break;
        }
        else if (P.at(j) != i && j == N-1){
          p.at(i) = i;
      }
    }
  }

  int count = 0;
  
  for (int j = 0; j < 55; j++) {
    if (count != 0){
      break;
    }
    for (int i = 0; i < 102; i++) {
      if (p.at(i) == X - j){
        cout << p.at(i) << endl;
        count ++;
        break;
      }
      else if (p.at(i) == X + j){
        cout << p.at(i) << endl;
        count ++;
        break;
      }
    }
  }
}