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() {
  int x, n;
  cin >> x, cin >> n;
  int p[n];
  for (int i = 0; i < n; i++) {
    cin >> p[i]; 
  }
  int a = 0;
  for (int j = 0; j < 100; j++) {
  int cnt = 0;
    for (int i = 0; i < n; i++) {
      if(p[i] == x + a) {
        a++;
        cnt = 1; 
      }
    }
  if (cnt == 0) break;
  }

  
  int b = 0;
  for (int j = 0; j < 100; j++) {
  int cnt = 0;
    for (int i = 0; i < n; i++) {
      if(p[i] == x - b) {
        b++;
        cnt = 1; 
      }
    }
    if (cnt == 0) break;
    }
  if (a < b) {
    cout << x + a << endl;  
  } else {
    cout << x - b << endl;  
  }
  return 0;
}