#include<bits/stdc++.h>
using namespace std;
 
const int MaxX=1e2;
const int MaxP=1e2;
bool IsOn[MaxP+2]={};
int main(){
	int X, N, p;
	scanf("%d %d\n",&X,&N);
	while(N-->0)
		scanf("%d",&p), IsOn[p]=1;
	int Grt, Lss;
	for(Grt=X;IsOn[Grt];Grt++);
	for(Lss=X;IsOn[Lss];Lss--);
	printf("%d\n",(Grt-X<X-Lss)? Grt: Lss);
}
### 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>
using namespace std;
 
const int MaxX=1e2;
const int MaxP=1e2;
bool IsOn[MaxP+2]={};
int main(){
	int X, N, p;
	scanf("%d %d\n",&X,&N);
	while(N-->0)
		scanf("%d",&p), IsOn[p]=1;
	int Grt, Lss;
	for(Grt=X;IsOn[Grt];Grt++);
	for(Lss=X;IsOn[Lss];Lss--);
	printf("%d\n",(Grt-X<X-Lss)? Grt: Lss);
}
```