#include <iostream>
#include <string>
#include <utility>
#include <map>
#define ll long long
using namespace std;

int main() {
	int N;
	string S;
	cin >> N >> S;
	map<pair<string, string>, int> mp;
	for(int i=0; i<(1<<N); ++i){
		string x="", y="";
		for(int d=0; d<N; ++d){
			if(i&(1<<d)) x += S[d];
			else y += S[d];
		}
		++mp[make_pair(x, y)];
	}
	ll ans=0;
	for(int i=0; i<(1<<N); ++i){
		string x="", y="";
		for(int d=0; d<N; ++d){
			if(i&(1<<d)) x += S[2*N-1-d];
			else y += S[2*N-1-d];
		}
		ans += mp[make_pair(x, y)];
	}
	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 <iostream>
#include <string>
#include <utility>
#include <map>
#define ll long long
using namespace std;

int main() {
	int N;
	string S;
	cin >> N >> S;
	map<pair<string, string>, int> mp;
	for(int i=0; i<(1<<N); ++i){
		string x="", y="";
		for(int d=0; d<N; ++d){
			if(i&(1<<d)) x += S[d];
			else y += S[d];
		}
		++mp[make_pair(x, y)];
	}
	ll ans=0;
	for(int i=0; i<(1<<N); ++i){
		string x="", y="";
		for(int d=0; d<N; ++d){
			if(i&(1<<d)) x += S[2*N-1-d];
			else y += S[2*N-1-d];
		}
		ans += mp[make_pair(x, y)];
	}
	cout << ans << endl;
	return 0;
}
```