#include <bits/stdc++.h>

const int N=210;
int bel[N],a[N],b[N],n;
bool G[N][N],vis[N];
std::map<int,int> M;
 
bool isprime(int n) {
	if(n==1) return 0;
	for(int i=2;i*i<=n;i++)
		if(n%i==0) return 0;
	return 1;
}
 
bool DFS(int i) {
	for(int j=1;j<=n;j++)
		if(!vis[j]&&G[i][j]) {
			vis[j]=1;
			if(!bel[j]||DFS(bel[j])) { bel[j]=i;return 1; }
		}
	return 0;
}
 
int main() {
	std::cin>>n;
	for(int i=1,x;i<=n;i++) std::cin>>x,M[x]^=1,M[x+1]^=1;
	for(auto x:M)
		if(x.second) (x.first&1?a[++*a]:b[++*b])=x.first;
	int ans=0;
	for(int i=1;i<=*a;i++)
		for(int j=1;j<=*b;j++)
			if(isprime(std::abs(a[i]-b[j]))) G[i][j]=1;
	for(int i=1;i<=*a;i++) {
		memset(vis,0,sizeof(vis));
		ans+=DFS(i);
	}
	std::cout<<ans+(*a-ans)/2*2+(*b-ans)/2*2+(*a-ans)%2*3<<std::endl;
	return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

There are infinitely many cards, numbered 1, 2, 3, ... Initially, Cards x_1, x_2, ..., x_N are face up, and the others are face down.

Snuke can perform the following operation repeatedly:

* Select a prime p greater than or equal to 3. Then, select p consecutive cards and flip all of them.



Snuke's objective is to have all the cards face down. Find the minimum number of operations required to achieve the objective.

Constraints

* 1 ≤ N ≤ 100
* 1 ≤ x_1 < x_2 < ... < x_N ≤ 10^7

Input

Input is given from Standard Input in the following format:


N
x_1 x_2 ... x_N


Output

Print the minimum number of operations required to achieve the objective.

Examples

Input

2
4 5


Output

2


Input

9
1 2 3 4 5 6 7 8 9


Output

3


Input

2
1 10000000


Output

4

### Response

```cpp
#include <bits/stdc++.h>

const int N=210;
int bel[N],a[N],b[N],n;
bool G[N][N],vis[N];
std::map<int,int> M;
 
bool isprime(int n) {
	if(n==1) return 0;
	for(int i=2;i*i<=n;i++)
		if(n%i==0) return 0;
	return 1;
}
 
bool DFS(int i) {
	for(int j=1;j<=n;j++)
		if(!vis[j]&&G[i][j]) {
			vis[j]=1;
			if(!bel[j]||DFS(bel[j])) { bel[j]=i;return 1; }
		}
	return 0;
}
 
int main() {
	std::cin>>n;
	for(int i=1,x;i<=n;i++) std::cin>>x,M[x]^=1,M[x+1]^=1;
	for(auto x:M)
		if(x.second) (x.first&1?a[++*a]:b[++*b])=x.first;
	int ans=0;
	for(int i=1;i<=*a;i++)
		for(int j=1;j<=*b;j++)
			if(isprime(std::abs(a[i]-b[j]))) G[i][j]=1;
	for(int i=1;i<=*a;i++) {
		memset(vis,0,sizeof(vis));
		ans+=DFS(i);
	}
	std::cout<<ans+(*a-ans)/2*2+(*b-ans)/2*2+(*a-ans)%2*3<<std::endl;
	return 0;
}
```