#include <bits/stdc++.h>
using namespace std;
using Int = long long;
pair<string, string> func(const string &s, int x)
{
    pair<string, string> P;
    for (int i = 0; i < s.size(); i++) {
        if (x & 1) {
            P.first += s[i];
        } else {
            P.second += s[i];
        }
        x >>= 1;
    }
    return P;
}
int main()
{
    int N; cin >> N;
    string S; cin >> S;
    string s1 = S.substr(0, N), s2 = S.substr(N, N);
    reverse(begin(s2), end(s2));
    map<pair<string, string>, int> M;
    Int ans = 0;
    for (int i = 0; i < (1 << N); i++) {
        M[func(s1, i)]++;
    }
    for (int i = 0; i < (1 << N); i++) {
        ans += M[func(s2, i)];
    }
    cout << ans << '\n';
    return 0;
}

### Prompt

Create a solution in cpp for 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;
using Int = long long;
pair<string, string> func(const string &s, int x)
{
    pair<string, string> P;
    for (int i = 0; i < s.size(); i++) {
        if (x & 1) {
            P.first += s[i];
        } else {
            P.second += s[i];
        }
        x >>= 1;
    }
    return P;
}
int main()
{
    int N; cin >> N;
    string S; cin >> S;
    string s1 = S.substr(0, N), s2 = S.substr(N, N);
    reverse(begin(s2), end(s2));
    map<pair<string, string>, int> M;
    Int ans = 0;
    for (int i = 0; i < (1 << N); i++) {
        M[func(s1, i)]++;
    }
    for (int i = 0; i < (1 << N); i++) {
        ans += M[func(s2, i)];
    }
    cout << ans << '\n';
    return 0;
}

```