#include <iostream>
#include <map>

using namespace std;
#define REP(i,n) for (int i = 0; i < n; ++i)

int N;
string S;
map<pair<string, string>, int> m;

int main() {
    cin >> N >> S;
    int all = (1<<N) - 1;
    REP (i, 1<<N) {
        string s, t;
        REP (j, N) {
            if (i>>(N-j-1) & 1) s += S[j];
            else t += S[j];
        }
        m[make_pair(s, t)]++;
    }

    long long ans = 0;
    REP (i, 1<<N) {
        string s, t;
        REP (j, N) {
            if (i>>(N-j-1) & 1) s += S[2*N-1-j];
            else t += S[2*N-1-j];
        }
        ans += m[make_pair(s, t)];
    }

    cout << ans << endl;
}
### Prompt

Please formulate 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 <iostream>
#include <map>

using namespace std;
#define REP(i,n) for (int i = 0; i < n; ++i)

int N;
string S;
map<pair<string, string>, int> m;

int main() {
    cin >> N >> S;
    int all = (1<<N) - 1;
    REP (i, 1<<N) {
        string s, t;
        REP (j, N) {
            if (i>>(N-j-1) & 1) s += S[j];
            else t += S[j];
        }
        m[make_pair(s, t)]++;
    }

    long long ans = 0;
    REP (i, 1<<N) {
        string s, t;
        REP (j, N) {
            if (i>>(N-j-1) & 1) s += S[2*N-1-j];
            else t += S[2*N-1-j];
        }
        ans += m[make_pair(s, t)];
    }

    cout << ans << endl;
}
```