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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<stdio.h>
//#include <bits/stdc++.h>
#include<vector>
#include<float.h>
#include<iomanip>
#include<algorithm>
#include<string>
#include<cstring>
#include<math.h>
#include<cmath>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include<cassert>
#include<cmath>
#include<cstdint>

#define INF 1e9
#define rep(i,n)for(int i=0;(i)<(int)(n);i++)
#define REP(i,a,b)for(int i=(int)(a);(i)<=(int)(b);i++)
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define vec(type,n) vector<type>(n)
#define vvec(m,n) vector<vector<int>> (int(m),vector<int>(n))
#define ALL(a)  (a).begin(),(a).end()

using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
using P = pair<char,int>;

const ll MOD = INF + 7;


int main(){
   int x, n;
   cin >> x >> n;
   if(n == 0){
      cout << x << endl;
      return 0;
   }

   vector<int> p(n);
   rep(i, n)cin >> p[i];
   sort(ALL(p));
   rep(i, 10000000){
      int now1 = x + i, now2 = x - i;
      if(now1 != *lower_bound(ALL(p), now1)){
         cout << now1 << endl;
         return 0;
      }
      if(now2 != *lower_bound(ALL(p), now2)){
         cout << now2 << endl;
         return 0;
      }
   }
}