#include<bits/stdc++.h>
using namespace std;
int n;
long long dp[21][21];
char s[101];
int main()
{
	cin>>n>>s;
	long long ans=0;
	for(int i=0;i<(1<<n);i++)
	{
		vector<int> red,blue;
		for(int j=0;j<n;j++)
			if(i&(1<<j))red.push_back(j);
			else blue.push_back(j);
		memset(dp,0,sizeof dp);
		int bluecnt=__builtin_popcount(i),redcnt=n-bluecnt;
		dp[0][0]=1;
		for(int j=0;j<n;j++)
			for(int k=0;k<=j&&k<=bluecnt;k++)
			{
				if(j-k<redcnt&&s[2*n-j-1]==s[blue[j-k]])dp[j+1][k]+=dp[j][k];
				if(k<bluecnt&&s[2*n-j-1]==s[red[k]])dp[j+1][k+1]+=dp[j][k];
			}
		ans+=dp[n][bluecnt];
	}
	cout<<ans<<endl;
	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<bits/stdc++.h>
using namespace std;
int n;
long long dp[21][21];
char s[101];
int main()
{
	cin>>n>>s;
	long long ans=0;
	for(int i=0;i<(1<<n);i++)
	{
		vector<int> red,blue;
		for(int j=0;j<n;j++)
			if(i&(1<<j))red.push_back(j);
			else blue.push_back(j);
		memset(dp,0,sizeof dp);
		int bluecnt=__builtin_popcount(i),redcnt=n-bluecnt;
		dp[0][0]=1;
		for(int j=0;j<n;j++)
			for(int k=0;k<=j&&k<=bluecnt;k++)
			{
				if(j-k<redcnt&&s[2*n-j-1]==s[blue[j-k]])dp[j+1][k]+=dp[j][k];
				if(k<bluecnt&&s[2*n-j-1]==s[red[k]])dp[j+1][k+1]+=dp[j][k];
			}
		ans+=dp[n][bluecnt];
	}
	cout<<ans<<endl;
	return 0;
}

```