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>
#define rep2(i,x,y) for (int i = x; i <= (y); ++i)
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
int INF = 100100101;

int vector_finder(std::vector<int> vec, int number) {
  auto itr = std::find(vec.begin(), vec.end(), number);
  size_t index = std::distance( vec.begin(), itr );
  if (index != vec.size()) { // 発見できたとき
    return 1;
  }
  else { // 発見できなかったとき
    return 0;
  }
}

int vector_index(vector<int> a,int target){
  vector<int>::iterator itr;
  itr = find(a.begin(), a.end(), target);
//  if(itr == a.end()) cout << "Not found" << endl;
  int ret = distance(a.begin(), itr);
  return ret;
}

int main()
{
  ios::sync_with_stdio(false);  cin.tie(0); cout.tie(0);
  int x, n;
  cin >> x >> n;
  vector<int> a(n);
  if(n == 0) {
    cout << x << endl;
    return 0;
  }
  cin>>a[0];
  int mini = abs(x - a[0]);
  int min_val = INF;
  for(int i = 1; i < n;i++){
    cin>>a[i];
    if(mini > abs(x - a[i])) {
      mini = abs(x - a[i]);
      min_val = a[i];
    }    
  }
  if(n == 1 && a[0] == x) {
   	cout << x << endl;
    return 0;
  }
  sort(a.begin(),a.end());
  int index = vector_index(a, min_val);
  int ansa = INF, ansb=INF;
  for(int i = 0; i + a[index] <= 100; i++){
    if(!vector_finder(a, i + a[index])) {
      ansa = i + a[index];
      cout << ansa << endl;
      break;
    }
  }
  for(int i = a[index]; i >= a[0]; i--){
   	 if(!vector_finder(a, i)) {
       ansb = i;
       cout << ansb << endl;
       break;
     }
  }
  cout << min(ansa, ansb) << endl;
  return 0;
}