#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define LL long long
#define st first
#define nd second
#define endl '\n'
using namespace std;

LL ans;

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	string s,a,b;
	cin>>n>>s;
	map<pair<string,string>,int > m;
	for(int i=0;i<(1<<n);++i) {
		a=b="";
		for(int j=0;j<n;++j)
			if(i&(1<<j)) {
				a+=s[j];
			}
			else b+=s[j];
		m[mp(a,b)]++;
	}
	for(int i=0;i<(1<<n);++i) {
		a=b="";
		for(int j=n-1;j>=0;--j)
			if(i&(1<<j)) {
				b+=s[j+n];
			}
			else a+=s[j+n];
		ans+=m[mp(a,b)];
	}
	cout<<ans<<endl;

}
### 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>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define LL long long
#define st first
#define nd second
#define endl '\n'
using namespace std;

LL ans;

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	string s,a,b;
	cin>>n>>s;
	map<pair<string,string>,int > m;
	for(int i=0;i<(1<<n);++i) {
		a=b="";
		for(int j=0;j<n;++j)
			if(i&(1<<j)) {
				a+=s[j];
			}
			else b+=s[j];
		m[mp(a,b)]++;
	}
	for(int i=0;i<(1<<n);++i) {
		a=b="";
		for(int j=n-1;j>=0;--j)
			if(i&(1<<j)) {
				b+=s[j+n];
			}
			else a+=s[j+n];
		ans+=m[mp(a,b)];
	}
	cout<<ans<<endl;

}
```