#include <bits/stdc++.h>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 10000000000000000
bool f = false;
map<pair<string,string>,int> get(string S){
	int N = S.size();
	map<pair<string,string>,int> mp;
	for(int i=0;i<(1<<N);i++){
		string s="",t="";
		for(int j=0;j<N;j++){
			if((i>>j)&1)s+=S[j];
			else t+=S[j];
		}
		reverse(t.begin(),t.end());
		if(f)swap(s,t);
		mp[make_pair(s,t)]++;
	}
	f=true;
	return mp;
}

int main(){
	
	int N;
	cin>>N;
	
	string S;
	cin>>S;
	
	map<pair<string,string>,int> mp0 = get(S.substr(0,N)),mp1 = get(S.substr(N));
	
	long long ans = 0;

	for(auto a:mp0){
		ans += (long long)a.second * mp1[a.first];
	}
	cout<<ans<<endl;
	
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 10000000000000000
bool f = false;
map<pair<string,string>,int> get(string S){
	int N = S.size();
	map<pair<string,string>,int> mp;
	for(int i=0;i<(1<<N);i++){
		string s="",t="";
		for(int j=0;j<N;j++){
			if((i>>j)&1)s+=S[j];
			else t+=S[j];
		}
		reverse(t.begin(),t.end());
		if(f)swap(s,t);
		mp[make_pair(s,t)]++;
	}
	f=true;
	return mp;
}

int main(){
	
	int N;
	cin>>N;
	
	string S;
	cin>>S;
	
	map<pair<string,string>,int> mp0 = get(S.substr(0,N)),mp1 = get(S.substr(N));
	
	long long ans = 0;

	for(auto a:mp0){
		ans += (long long)a.second * mp1[a.first];
	}
	cout<<ans<<endl;
	
	return 0;
}
```