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<iostream> 
#include<stdio.h>
#include<string>
#include<iomanip> 
#include<vector> 
#include<map> 
#include<set> 
#include<algorithm> 
#include<numeric> 
#include<limits> 
#include<bitset> 
#include<functional> 
#include<type_traits> 
#include<queue> 
#include<stack> 
#include<array> 
#include<random> 
#include<utility> 
#include<cstdlib> 
#include<ctime>
//#define _LIBCPP_DEBUG 0
//#define _GLIBCXX_DEBUG
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define all(a) (a).begin(),(a),end()
#define rep(i,n) for (ll i = 0; i < (n); ++i)	
//setprecision(16)

void outi(int x) { cout << x << endl; }
void outl(ll x) { cout << x << endl; }
void outs(string x) { cout << x << endl; }
ll max(ll a, ll b) {
	ll max;
	if (a > b) max = a;
	else max = b;
	return max;
}
ll min(ll a, ll b) {
	ll min;
	if (a < b) min = a;
	else min = b;
	return min;
}
//---------------------------------------------------------------

int main() {

	int x, n;
	cin >> x >> n;
	int p[110];

	rep(i, n) cin >> p[i];
	int res[110] = { 0 };
	if (n == 0) {
		cout << x << endl;
		return 0;
	}
	rep(i, n) {
		res[p[i]-1] += 1;
	}

	int ans = 0;
	int k = 0;
	rep(i, 101) {
		
		if (res[i] == 0 && abs(k - x) > abs(i + 1 - x)) {
			k = i;
			ans = i + 1;

		}
	}

	outi(ans);
}