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;
using ll = long long;
using P = pair<int,int>;

int main() {
   	int x, N;
	cin >> x >> N;
  
    if (N==0) {
      cout << x << endl;
      return 0;
    }
  
  	vector<int> p(101, 1);
  	rep(i, N) {
      int ibuff;
      cin >> ibuff;
      p.at(ibuff) = 0;
    }
  	
    int i1=-100, i2=100;
	for (int i=x; i>-100; --i) {
      if (p.at(i)==1) {
        i1=i;
        break;
      }
    }
      
  	for (int i=x; i<=100; ++i) {
      if (p.at(i)==1) {
        i2=i;
        break;
      }
    }

    if (x-i1 <= i2-x) {
      cout << i1 << endl;
    } else {
      cout << i2 << endl;
    }
    return 0;
}