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(void) {
  int x, n;
  cin >> x >> n;
  vector<int> ps(n);
  for (int &p : ps) cin >> p;

  // n == 0場合もあるる
  if (n == 0) {
    cout << x << endl;
    return 0;
  }

  int lower = -1;
  int upper = 0;
  while (true) {
    if (lower < -100 && upper > 100) break;

    auto result0 = find(ps.begin(), ps.end(), min(101, x + upper));
    if (result0 == ps.end()) {
      cout << min(101, x + upper) << endl;
      return 0;
    }
    upper += 1;
    
    auto result1 = find(ps.begin(), ps.end(), max(0, x + lower));
    if (result1 == ps.end()) {
      cout << max(0, x + lower) << endl;
      return 0;
    }
    lower -= 1;

  }

  return 0;
}