#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define Would
#define you
#define please

int main() {
	
	//解説PDFを見ました(迫真)
	//速度の様子うかがい

	int N;
	string S;
	cin >> N >> S;

	map<string, int> GOD;

	rep(i, 1 << N) {
		string blue;
		string red;
		rep(ii, N) {
			if (i >> ii & 1) {
				blue += S[ii];
			}
			else {
				red += S[ii];
			}
		}
		GOD[blue + "," + red]++;
	}
	long long kotae = 0;
	rep(i, 1 << N) {
		string blue2;
		string red2;
		rep(ii, N) {
			if (i >> ii & 1) {
				blue2 += S[2 * N - ii - 1];
			}
			else {
				red2 += S[2 * N - ii - 1];
			}
		}
		kotae += GOD[blue2 + "," + red2];
	}
	cout << kotae << "\n";
	Would you please return 0;

}

### Prompt

Generate 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>
using namespace std;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define Would
#define you
#define please

int main() {
	
	//解説PDFを見ました(迫真)
	//速度の様子うかがい

	int N;
	string S;
	cin >> N >> S;

	map<string, int> GOD;

	rep(i, 1 << N) {
		string blue;
		string red;
		rep(ii, N) {
			if (i >> ii & 1) {
				blue += S[ii];
			}
			else {
				red += S[ii];
			}
		}
		GOD[blue + "," + red]++;
	}
	long long kotae = 0;
	rep(i, 1 << N) {
		string blue2;
		string red2;
		rep(ii, N) {
			if (i >> ii & 1) {
				blue2 += S[2 * N - ii - 1];
			}
			else {
				red2 += S[2 * N - ii - 1];
			}
		}
		kotae += GOD[blue2 + "," + red2];
	}
	cout << kotae << "\n";
	Would you please return 0;

}

```