#include<cstdio>
#include<algorithm>
#include<map>
#include<string>
#define pss pair<string,string>
using namespace std;
map<pss,int>Map;
int n;
char p[40];
void Go1(int pv, int e, string &a, string &b) {
	if (pv == e) {
		Map[pss(a, b)]++;
		return;
	}
	a += p[pv];
	Go1(pv + 1, e, a, b);
	a.pop_back();
	b += p[pv];
	Go1(pv + 1, e, a, b);
	b.pop_back();
}
long long res = 0;
void Go2(int pv, int e, string &a, string &b) {
	if (pv == e) {
		res += Map[pss(b, a)];
		return;
	}
	a += p[pv];
	Go2(pv - 1, e, a, b);
	a.pop_back();
	b += p[pv];
	Go2(pv - 1, e, a, b);
	b.pop_back();
}
int main() {
	int i;
	scanf("%d", &n);
	scanf("%s", p);
	string t1="", t2="";
	Go1(0, n, t1, t2);
	Go2(2 * n - 1, n - 1, t1, t2);
	printf("%lld\n", res);
}
### Prompt

Your challenge is to write a CPP solution 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<cstdio>
#include<algorithm>
#include<map>
#include<string>
#define pss pair<string,string>
using namespace std;
map<pss,int>Map;
int n;
char p[40];
void Go1(int pv, int e, string &a, string &b) {
	if (pv == e) {
		Map[pss(a, b)]++;
		return;
	}
	a += p[pv];
	Go1(pv + 1, e, a, b);
	a.pop_back();
	b += p[pv];
	Go1(pv + 1, e, a, b);
	b.pop_back();
}
long long res = 0;
void Go2(int pv, int e, string &a, string &b) {
	if (pv == e) {
		res += Map[pss(b, a)];
		return;
	}
	a += p[pv];
	Go2(pv - 1, e, a, b);
	a.pop_back();
	b += p[pv];
	Go2(pv - 1, e, a, b);
	b.pop_back();
}
int main() {
	int i;
	scanf("%d", &n);
	scanf("%s", p);
	string t1="", t2="";
	Go1(0, n, t1, t2);
	Go2(2 * n - 1, n - 1, t1, t2);
	printf("%lld\n", res);
}
```