#include<bits/stdc++.h>
using namespace std;

int main() {
    int64_t N;
    string S;
    cin >> N >> S;

    reverse(S.begin() + N, S.end());

    map<pair<string, string>, int64_t> mp[2];
    for (int64_t i = 0; i < (1LL << N); i++) {
        for (int64_t j = 0; j < 2; j++) {
            string str1, str2;
            for (int64_t k = 0; k < N; k++) {
                if (i & (1LL << k)) {
                    str1 += S[k + N * j];
                } else {
                    str2 += S[k + N * j];
                }
            }

            mp[j][make_pair(str1, str2)]++;
        }
    }

    int64_t ans = 0;
    for (const auto& p : mp[0]) {
        ans += mp[0][p.first] * mp[1][p.first];
    }
    cout << ans << endl;
}
### 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>
using namespace std;

int main() {
    int64_t N;
    string S;
    cin >> N >> S;

    reverse(S.begin() + N, S.end());

    map<pair<string, string>, int64_t> mp[2];
    for (int64_t i = 0; i < (1LL << N); i++) {
        for (int64_t j = 0; j < 2; j++) {
            string str1, str2;
            for (int64_t k = 0; k < N; k++) {
                if (i & (1LL << k)) {
                    str1 += S[k + N * j];
                } else {
                    str2 += S[k + N * j];
                }
            }

            mp[j][make_pair(str1, str2)]++;
        }
    }

    int64_t ans = 0;
    for (const auto& p : mp[0]) {
        ans += mp[0][p.first] * mp[1][p.first];
    }
    cout << ans << endl;
}
```