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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int main(){
  int X, N;
  cin >> X >> N;
  vector<int> a(N);
  if(N == 0) cout << X << endl;
  else{
    for(int i = 0; i < N; i++){
      cin >> a.at(i);
    }
    sort(a.begin(),a.end());
    vector<int> b;
    for(int i = 0; i <= 101; i++){
      for(int j = 0; j < a.size(); j++){
        if(a.at(j) == i ){
          break;
        }
        else if(a.at(j) > i){
          b.push_back(i);
          break;
        }
      }
    }
    int v;
    map<int, int> m;
    for(int i = 0; i < b.size(); i++){
      v = X - b.at(i);
      if(v < 0) v = v * (-1);
      m[v] = b.at(i);
    }
    vector<int> c(4);
    int count = 0;
    for(auto p : m){
      c[count] = p.first;
      c[count + 1] = p.second;
      count = count + 2;
    }
    if(c.at(0) == c.at(2)){
      cout << min(c.at(1), c.at(3)) << endl;
    }
    else cout << c.at(1) << endl;
  }
  return 0;
}