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>
using namespace std;

#define REP(i, x) for (int i = 0; i < (int)(x); i++)
#define REPS(i, x) for (int i = 1; i <= (int)(x); i++)
#define RREP(i, x) for (int i = ((int)(x)-1); i >= 0; i--)
#define RREPS(i, x) for (int i = ((int)(x)); i > 0; i--)
#define FOR(i, a, b) for (int i = a; i <= (int)(b); i++)
#define FORD(i, a, b) for (int i = a; i >= (int)(b); i--)

#define SIZE_OF_ARRAY(array) (sizeof(array) / sizeof(array[0]))
using Lint = long long;
#define INF 1000000000000
#define MOD 1000000007 //10^9+7:合同式の法

#define ALL(x) (x).begin(), (x).end()
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))

#define PB push_back
#define MP make_pair
using P=pair<int, int>;
using V=vector<int>;
using M=map<int, int>;
using S=set<int>;

#define COUT(x) cout << (x) << endl
#define SCOUT(x) cout << (x) << " "

#define DUMP(x) cout << #x << " = " << (x) << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
template <class T>bool chmax(T &a, const T &b){if (a < b){a = b;return 1;}return 0;}
template <class T>bool chmin(T &a, const T &b){if (b < a){a = b;return 1;}return 0;}
/* vector<T>のイテレータはvector<T>::iterator型
vector<int>::iterator it = vc.begin();  // begin()はvectorの最初の要素を指すイテレータを返す。
vector<int>::iterator endIt = vc.end(); // end()はvectorの終端位置を指すイテレータを返す。
while (it != endIt)
{
  cout << *it << '\n'; // イテレータの指す要素は、ポインタのように*演算子でアクセスします。
  it++;                // インクリメントすることで次の要素を指します。
}
}*/
int main()
{
  
  int x,n,aa=1,f=-1;
  cin>>x>>n;
  V a(n);
  REP(i,n)cin>>a[i];
  int b=100;
  while(b){
    REP(i,n){
  		if(x==a[i])
  		{x-=aa;
  		aa++;
  		aa*=f;
  		f*=(-1);
  		}
	 }
    b--;
  }
  if(n==0)COUT(x);
  else COUT(x);
}