#include <bits/stdc++.h>
#define Int int64_t

using namespace std;

int main() {
	int N;
	string s;
	cin >> N >> s;

	string sL = s.substr(0, N), sR = s.substr(N);
	reverse(sR.begin(), sR.end());
	map<pair<string, string>, Int> L, R;
	auto f = [&](string S, map<pair<string, string>, Int>& mp) -> void {
		for (int i = 0; i < (1<<N); ++i) {
			string X = "", Y = "";
			for (int j = 0; j < N; ++j) {
				if ((i>>j) & 1) { X.push_back(S[j]); }
				else { Y.push_back(S[j]); }
			}
			mp[{ X, Y }] += 1;
		}
	};

	f(sL, L);
	f(sR, R);
	Int ans = 0;
	for (auto it = L.begin(); it != L.end(); ++it) {
		auto p = it->first;
		if (!R.count(p)) { continue; }
		ans += it->second * R[p];
	}
	cout << ans << "\n";

	return 0;
}

### 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 <bits/stdc++.h>
#define Int int64_t

using namespace std;

int main() {
	int N;
	string s;
	cin >> N >> s;

	string sL = s.substr(0, N), sR = s.substr(N);
	reverse(sR.begin(), sR.end());
	map<pair<string, string>, Int> L, R;
	auto f = [&](string S, map<pair<string, string>, Int>& mp) -> void {
		for (int i = 0; i < (1<<N); ++i) {
			string X = "", Y = "";
			for (int j = 0; j < N; ++j) {
				if ((i>>j) & 1) { X.push_back(S[j]); }
				else { Y.push_back(S[j]); }
			}
			mp[{ X, Y }] += 1;
		}
	};

	f(sL, L);
	f(sR, R);
	Int ans = 0;
	for (auto it = L.begin(); it != L.end(); ++it) {
		auto p = it->first;
		if (!R.count(p)) { continue; }
		ans += it->second * R[p];
	}
	cout << ans << "\n";

	return 0;
}

```