#include <bits/stdc++.h>
using namespace std;

int main(int argc, char const *argv[])
{
	int n;
	string s;
	cin>>n>>s;
	map<pair<string,string>,int> mp;
	for (int i = 0; i < (1<<n); ++i)
	{
		string l,r;
		for (int j = 0; j < n; ++j)
		{
			if ((i>>j)&1)
			{
				l.push_back(s[j]);
			}
			else
			{
				r.push_back(s[j]);
			}
		}
		++mp[make_pair(l,r)];
	}
	long long ans=0LL;
	for (int i = 0; i < (1<<n); ++i)
	{
		string l,r;
		for (int j = 0; j < n; ++j)
		{
			if ((i>>j)&1)
			{
				r.push_back(s[j+n]);
			}
			else
			{
				l.push_back(s[j+n]);
			}
		}
		reverse(l.begin(), l.end());
		reverse(r.begin(), r.end());
		ans+=mp[make_pair(l,r)];
	}
	cout<<ans<<endl;
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 main(int argc, char const *argv[])
{
	int n;
	string s;
	cin>>n>>s;
	map<pair<string,string>,int> mp;
	for (int i = 0; i < (1<<n); ++i)
	{
		string l,r;
		for (int j = 0; j < n; ++j)
		{
			if ((i>>j)&1)
			{
				l.push_back(s[j]);
			}
			else
			{
				r.push_back(s[j]);
			}
		}
		++mp[make_pair(l,r)];
	}
	long long ans=0LL;
	for (int i = 0; i < (1<<n); ++i)
	{
		string l,r;
		for (int j = 0; j < n; ++j)
		{
			if ((i>>j)&1)
			{
				r.push_back(s[j+n]);
			}
			else
			{
				l.push_back(s[j+n]);
			}
		}
		reverse(l.begin(), l.end());
		reverse(r.begin(), r.end());
		ans+=mp[make_pair(l,r)];
	}
	cout<<ans<<endl;
	return 0;
}

```