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>
#include <iomanip>
#include <algorithm>
#define ll long long
#include <unordered_map>
using namespace std;
 
int main() {
  ll X, N, ans, min = 114514;
  cin >> X >> N;
  bool ansX = 1, continuefr = 0;
  vector<ll> p(N);
  for(ll i = 0; i < N; i++) {
    cin >> p.at(i);
    if(p.at(i) == X) ansX = 0;
  }
  if(ansX) cout << X << endl;
  else {
    for(ll i = 0; i <= 101; i++) {
      continuefr = 0;
      for(ll j = 0; j < N; j++) {
        if(i == p.at(j)) {
          continuefr = 1;
          break;
        }
      }
      if(continuefr == 0) {
        if(abs(X - i) < min) {
          min = abs(X - i);
          ans = i;
        }
      }
    }
  cout << ans << endl;  
  }
}