#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <math.h>
#include <iostream>
#include <map>
using namespace std;

typedef long long ll;
const int MAXN = 36 + 10;
int N;
char S[MAXN];
map<pair<string, string>, ll> mp;
ll ans;

int main() {
	register int s, i;
	scanf("%d%s", &N, S);
	for(s = 0; s < (1 << N); ++s) {
		string s1, s2;
		for(i = 0; i < N; ++i)
			if(s >> i & 1)
				s1.push_back(S[i]);
			else
				s2.push_back(S[i]);
		++mp[make_pair(s1, s2)];
	}
	for(s = 0; s < (1 << N); ++s) {
		string s1, s2;
		for(i = N - 1; ~i; --i)
			if(s >> i & 1)
				s2.push_back(S[N + i]);
			else
				s1.push_back(S[N + i]);
		ans += mp[make_pair(s1, s2)];
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <cstdlib>
#include <cstring>
#include <algorithm>
#include <math.h>
#include <iostream>
#include <map>
using namespace std;

typedef long long ll;
const int MAXN = 36 + 10;
int N;
char S[MAXN];
map<pair<string, string>, ll> mp;
ll ans;

int main() {
	register int s, i;
	scanf("%d%s", &N, S);
	for(s = 0; s < (1 << N); ++s) {
		string s1, s2;
		for(i = 0; i < N; ++i)
			if(s >> i & 1)
				s1.push_back(S[i]);
			else
				s2.push_back(S[i]);
		++mp[make_pair(s1, s2)];
	}
	for(s = 0; s < (1 << N); ++s) {
		string s1, s2;
		for(i = N - 1; ~i; --i)
			if(s >> i & 1)
				s2.push_back(S[N + i]);
			else
				s1.push_back(S[N + i]);
		ans += mp[make_pair(s1, s2)];
	}
	printf("%lld\n", ans);
	return 0;
}
```