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 rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

int main() {
  int X, N;
  cin >> X >> N;
  set<int> p;
  rep(i,N){
    int tmpP;
    cin >> tmpP;
    p.insert(tmpP);
  }

  if(p.size() == 0){
    cout << X << endl;
    return 0;
  }

  int minNum = -1;
  int minDiff = 9999999;
  for (int i = -1; i < 101; i++){
    if(p.count(i)){
      continue;
    }
    int diff = X - i;
    if (diff < 0) diff *= -1;

    if(diff < minDiff){
       minDiff = diff;
       minNum = i;
    }
    if(minDiff == diff && i < minNum){
       minNum = i;
    }
  }
  cout << minNum << endl;
}