#include <iostream>
#include <algorithm>
#include <string>
#include <unordered_map>

using namespace std;
using Map = unordered_map<string, long>;

size_t n;

void dfs(string src, Map &m, size_t i=0, string s="", string t="") {
    if (i == n){
        m[s + ' ' + t]++;
    } else {
        dfs(src, m, i+1, s+src[i], t);
        dfs(src, m, i+1, s, t+src[i]);
    }
}

int main() {
    string src;
    Map m1, m2;
    cin >> n >> src;
    string left  = src.substr(0, n),
           right = src.substr(n);
    reverse(right.begin(), right.end());
    dfs(left, m1);
    dfs(right, m2);
    long cnt = 0;
    for (auto mp:m1) 
        cnt += mp.second * m2[mp.first];
    cout << cnt << '\n';
    return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <algorithm>
#include <string>
#include <unordered_map>

using namespace std;
using Map = unordered_map<string, long>;

size_t n;

void dfs(string src, Map &m, size_t i=0, string s="", string t="") {
    if (i == n){
        m[s + ' ' + t]++;
    } else {
        dfs(src, m, i+1, s+src[i], t);
        dfs(src, m, i+1, s, t+src[i]);
    }
}

int main() {
    string src;
    Map m1, m2;
    cin >> n >> src;
    string left  = src.substr(0, n),
           right = src.substr(n);
    reverse(right.begin(), right.end());
    dfs(left, m1);
    dfs(right, m2);
    long cnt = 0;
    for (auto mp:m1) 
        cnt += mp.second * m2[mp.first];
    cout << cnt << '\n';
    return 0;
}
```