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;

int main() {

  int X, N;
  cin >> X >> N;
  
  if(N == 0) cout << X << endl;
  else {
  
    vector<int> p(N);
    for(int i = 0; i < N; i++) {
      cin >> p.at(i);
    }
    sort(p.begin(), p.end());
    for(int i = 0; i < N; i++) {
    }

    int s_flag = p.at(0)-1;
    int l_flag = p.at(N-1)+1;
  
    for(int i = 0; i < N-1; i++) {
      if(p.at(i)+1 < p.at(i+1)) {
        if(p.at(i) < X && X < p.at(i+1)) cout << X << endl;
        else if(X <= p.at(i) && i < N-1) l_flag = p.at(i)+1;
        else if(p.at(i+1) <= X) s_flag = p.at(i+1)-1;
      }
    }
  
    if(l_flag - X < X - s_flag) cout << l_flag << endl;
    else cout << s_flag << endl;
  }
         
}