#include <iostream>
#include <string>
#include <map>
#include <utility>
using namespace std;
typedef long long ll;
string S;
int N;
int main(){
	cin >> N >> S;
	map<pair<string,string>,ll> ml,mr;
	string L = S.substr(0,N),R = S.substr(N,N);
	for(int i=0;i<(1<<N);i++){
		string lr = "",lb = "",rr = "",rb = "";
		for(int j=0;j<N;j++){
			if(i & (1<<j)){
				lr += L[j];
				rr += R[N-1-j];
			}else{
				lb += L[j];
				rb += R[N-1-j];
			}
		}
//		cout << lr << " " << lb << endl;
		ml[make_pair(lr,lb)]++;
		mr[make_pair(rr,rb)]++;
	}
	ll ans = 0;
	for(auto x:ml){
		ans += x.second*mr[x.first];
	}
	cout << ans << endl;
}
### Prompt

Please create a solution in cpp to 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 <iostream>
#include <string>
#include <map>
#include <utility>
using namespace std;
typedef long long ll;
string S;
int N;
int main(){
	cin >> N >> S;
	map<pair<string,string>,ll> ml,mr;
	string L = S.substr(0,N),R = S.substr(N,N);
	for(int i=0;i<(1<<N);i++){
		string lr = "",lb = "",rr = "",rb = "";
		for(int j=0;j<N;j++){
			if(i & (1<<j)){
				lr += L[j];
				rr += R[N-1-j];
			}else{
				lb += L[j];
				rb += R[N-1-j];
			}
		}
//		cout << lr << " " << lb << endl;
		ml[make_pair(lr,lb)]++;
		mr[make_pair(rr,rb)]++;
	}
	ll ans = 0;
	for(auto x:ml){
		ans += x.second*mr[x.first];
	}
	cout << ans << endl;
}
```