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<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<algorithm>
#include<cmath>
#include<bitset>
 
#define Vsort(a) sort(a.begin(), a.end())
#define Vreverse(a) reverse(a.begin(), a.end())
#define Srep(n) for(int i = 0; i < (n); i++)
#define Lrep(i,a,n) for(int i = (a); i < (n); i++)
#define Brep(n) for(int bit = 0; bit < (1<<n); bit++)
#define rep2nd(n,m) Srep(n) Lrep(j,0,m)
#define vi vector<int>
#define vi64 vector<int64_t>
#define vvi vector<vector<int>>
#define vvi64 vector<vector<int64_t>>
 
using namespace std;

int main(){
  int x,n;
  cin >> x >> n;
  if(n == 0){
    cout << x << endl;
    return 0;
  }
  vi p(n),q;
  Srep(n) cin >> p[i];
  Vsort(p);
  int i = 0;
  Lrep(j,1,101){
    if(i > n - 1) i--;
    else if(p[i] == j) i++;
    else q.push_back(j);
  }
  int m = q.size(),min = 100,ans=0;
  Srep(m){
    if(abs(x - q[i]) < min){
      ans = q[i];
      min = abs(x - q[i]);
    }
  }
  cout << ans << endl;
}