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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair <int, int> P;

template <typename T> T gcd(T x, T y) { return( y ? gcd(y, x % y) : x); }
template <typename T> T lcm(T x, T y) { return( x / gcd(x, y) * y); }
template <typename T> void printVec(vector<T> &vec) {
  cout << "";
  for (auto it = vec.begin(); it != vec.end(); ++it) {
    cout << *it << " ";
  }
  cout << endl;
}

int main() {
  int x, n;
  cin >> x >> n;
  vector <int> p(n);

  for(int i = 0; i < n; i++) {
    cin >> p[i];
  }
  int v = INT_MAX;
  int ans = -1;
  sort(p.begin(), p.end());
  int point = 0;
  for(int i = 1; i <= 100; i++) {
    if (n != 0) {
      if (i != p[point]) {
        // cout << "i = " << i << endl;
        int d = abs(i - x);
        if (d < v) {
          v = d;
          ans = i;
        }
      } else {
        point = point + 1;
      }
    } else {
      int d = abs(i - x);
      if (d < v) {
        v = d;
        ans = i;
      }     
    }
  }
  cout << ans << endl;
	return 0;
}