#include<cstdio>
int p[1001],x,n;
int main(){
  scanf("%d%d",&x,&n);
  for(int i=1;i<=n;i++){
    int a;scanf("%d",&a),p[a]=1;
  }
  for(int i=0;i<=1001;i++){
    if(x-i>=0){
      if(!p[x-i])return printf("%d\n",x-i)&0;
    }
    if(!p[x+i])return printf("%d\n",x+i)&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<cstdio>
int p[1001],x,n;
int main(){
  scanf("%d%d",&x,&n);
  for(int i=1;i<=n;i++){
    int a;scanf("%d",&a),p[a]=1;
  }
  for(int i=0;i<=1001;i++){
    if(x-i>=0){
      if(!p[x-i])return printf("%d\n",x-i)&0;
    }
    if(!p[x+i])return printf("%d\n",x+i)&0;
  }
}
```