#include <bits/stdc++.h>
#define lol long long
using namespace std;
int main(){
  int n,x,p,ans=10000,a[200]={};
  cin>>x>>n;
  for(int i=0;i<n;i++){
    cin>>p;
    a[p]=1;
  }
  for(int i=0;i<=101;i++){
    if(a[i]==0 && abs(x-i)<abs(x-ans)) ans=i;
    if(abs(x-i)==abs(x-ans) && i<ans) ans=i;
  }
  cout<<ans<<"\n";
  return(0);
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

Examples

Input

6 5
4 7 10 6 5


Output

8


Input

10 5
4 7 10 6 5


Output

9


Input

100 0


Output

100

### Response

```cpp
#include <bits/stdc++.h>
#define lol long long
using namespace std;
int main(){
  int n,x,p,ans=10000,a[200]={};
  cin>>x>>n;
  for(int i=0;i<n;i++){
    cin>>p;
    a[p]=1;
  }
  for(int i=0;i<=101;i++){
    if(a[i]==0 && abs(x-i)<abs(x-ans)) ans=i;
    if(abs(x-i)==abs(x-ans) && i<ans) ans=i;
  }
  cout<<ans<<"\n";
  return(0);
}
```