#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long llu;
typedef pair<int, int> pii;
#define x first
#define y second
const int INF = 1000000000, MAXN = 18;

string str;
map<string, map<string, int> > mp;
int N;
ll res;

int main(){
	cin>>N>>str;
	for(int mask=0; mask<(1<<N); ++mask){
		string s1, s2;
		for(int j=0; j<N; ++j){
			if(mask & (1<<j)) s1 += str[N + j];
			else s2 += str[N + j];
		}
		reverse(s2.begin(), s2.end());
		mp[s1][s2]++;
	}
	for(int mask=0; mask<(1<<N); ++mask){
		string s1, s2;
		for(int j=0; j<N; ++j){
			if(mask & (1<<j)) s1 += str[j];
			else s2 += str[j];
		}
		reverse(s1.begin(), s1.end());
		res += mp[s1][s2];
	}
	printf("%lld\n", res);
	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;

typedef long long ll;
typedef unsigned long long llu;
typedef pair<int, int> pii;
#define x first
#define y second
const int INF = 1000000000, MAXN = 18;

string str;
map<string, map<string, int> > mp;
int N;
ll res;

int main(){
	cin>>N>>str;
	for(int mask=0; mask<(1<<N); ++mask){
		string s1, s2;
		for(int j=0; j<N; ++j){
			if(mask & (1<<j)) s1 += str[N + j];
			else s2 += str[N + j];
		}
		reverse(s2.begin(), s2.end());
		mp[s1][s2]++;
	}
	for(int mask=0; mask<(1<<N); ++mask){
		string s1, s2;
		for(int j=0; j<N; ++j){
			if(mask & (1<<j)) s1 += str[j];
			else s2 += str[j];
		}
		reverse(s1.begin(), s1.end());
		res += mp[s1][s2];
	}
	printf("%lld\n", res);
	return 0;
}

```