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>
#include <string>
using namespace std;
#define ll long long
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define rep2(i, b) for (int i = 0; i < b; i++)
#define repl(i, a, b) for (ll i = a; i <= b; i++)
#define ppi pair<int, int>
#define scl(n) scanf("%d", &n);
#define pcl(n) printf("%d\n", n)
#define pcl1(n) printf("%lld ", n)
#define FastRead                  \
    ios_base::sync_with_stdio(0); \
    cin.tie(0)
bool comp(ppi a, ppi b)
{
    return a.first < b.first;
}
int a[205];
vector<int>v;
int main()
{
    FastRead;
    int x, n,i = 1;
    cin >> x >> n;
	int a[n+6]={0};
    for(int i=1;i<=n;i++)
	{
		int p;
		cin>>p;
		a[p]=p;
	}
	if(n>0)
	{
		int z=100000,res=0;
		for(int i=1;i<=105;i++)
		{
			if(!a[i])
			{
				if((int)abs(x-i)<z)
				{
					z=(int)abs(x-i);
					res=i;
				}
			}
		}
		cout<<res<<endl;
	}
	else cout<<x<<endl;
    return 0;
}