#include<cstring>
#include<cstdio>
#include<map>
using namespace std;
const int MAXN=20;
const unsigned int HASH=103U,MOD=1000000007U;

map<pair<unsigned int,unsigned int>,int> cnt;
char str[MAXN*2];

int main()
{
	int n;
	scanf("%d%s",&n,str);
	for(int s=0;s<(1<<n);s++)
	{
		unsigned int a=0,b=0;
		for(int i=0;i<n;i++)
			if(s&(1<<i))
				a=(1LLU*a*HASH+str[i]-'a'+1)%MOD;
			else
				b=(1LLU*b*HASH+str[i]-'a'+1)%MOD;
		cnt[make_pair(a,b)]++;
	}
	
	long long ans=0;
	for(int s=0;s<(1<<n);s++)
	{
		unsigned int a=0,b=0;
		for(int i=0;i<n;i++)
			if(s&(1<<i))
				b=(1LLU*b*HASH+str[2*n-i-1]-'a'+1)%MOD;
			else
				a=(1LLU*a*HASH+str[2*n-i-1]-'a'+1)%MOD;
		ans+=cnt[make_pair(a,b)];
	}
	
	printf("%lld\n",ans);
	
	return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

You are given a string S of length 2N consisting of lowercase English letters.

There are 2^{2N} ways to color each character in S red or blue. Among these ways, how many satisfy the following condition?

* The string obtained by reading the characters painted red from left to right is equal to the string obtained by reading the characters painted blue from right to left.

Constraints

* 1 \leq N \leq 18
* The length of S is 2N.
* S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the number of ways to paint the string that satisfy the condition.

Examples

Input

4
cabaacba


Output

4


Input

11
mippiisssisssiipsspiim


Output

504


Input

4
abcdefgh


Output

0


Input

18
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa


Output

9075135300

### Response

```cpp
#include<cstring>
#include<cstdio>
#include<map>
using namespace std;
const int MAXN=20;
const unsigned int HASH=103U,MOD=1000000007U;

map<pair<unsigned int,unsigned int>,int> cnt;
char str[MAXN*2];

int main()
{
	int n;
	scanf("%d%s",&n,str);
	for(int s=0;s<(1<<n);s++)
	{
		unsigned int a=0,b=0;
		for(int i=0;i<n;i++)
			if(s&(1<<i))
				a=(1LLU*a*HASH+str[i]-'a'+1)%MOD;
			else
				b=(1LLU*b*HASH+str[i]-'a'+1)%MOD;
		cnt[make_pair(a,b)]++;
	}
	
	long long ans=0;
	for(int s=0;s<(1<<n);s++)
	{
		unsigned int a=0,b=0;
		for(int i=0;i<n;i++)
			if(s&(1<<i))
				b=(1LLU*b*HASH+str[2*n-i-1]-'a'+1)%MOD;
			else
				a=(1LLU*a*HASH+str[2*n-i-1]-'a'+1)%MOD;
		ans+=cnt[make_pair(a,b)];
	}
	
	printf("%lld\n",ans);
	
	return 0;
}

```