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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define mod 1000000007
#define dbg(n) cout<<#n<<' '<<n<<endl;
int main(){ 
	int x,n;
	cin>>x>>n;
	int a[102]={0};
	for(int i=0;i<n;i++){
		int y;
		cin>>y;
		a[y]=1;
	}	
	int mn=1e9,ind;
	for(int i=0;i<=101;i++){
		if(mn>abs(i-x)&&!a[i]){
			mn=abs(i-x);
			ind=i;
		}
	}
	cout<<ind<<endl;
}