#include <map>
#include <iostream>
using namespace std;
typedef pair<string,string> pa;
int main() {
    int n;
    long long ans = 0;
    string s;
    map<pa,int> mp;
    cin>>n>>s;
    for(int i = 0;i < 1 << n;i ++) {
        string a = "";
        string b = "";
        for(int j = 0;j < n;j ++) {
            if(i >> j & 1)a += s[j];
            else b += s[j];
        }
        mp[pa(a,b)] ++;
    }
    for(int i = 0;i < 1 << n;i ++) {
        string a = "";
        string b = "";
        for(int j = 0;j < n;j ++) {
            if(i >> j & 1)a += s[n * 2 - 1 - j];
            else b += s[n * 2 - 1 - j];
        }
        ans += mp[pa(a,b)];
    }
    cout<<ans;
}

### 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 <map>
#include <iostream>
using namespace std;
typedef pair<string,string> pa;
int main() {
    int n;
    long long ans = 0;
    string s;
    map<pa,int> mp;
    cin>>n>>s;
    for(int i = 0;i < 1 << n;i ++) {
        string a = "";
        string b = "";
        for(int j = 0;j < n;j ++) {
            if(i >> j & 1)a += s[j];
            else b += s[j];
        }
        mp[pa(a,b)] ++;
    }
    for(int i = 0;i < 1 << n;i ++) {
        string a = "";
        string b = "";
        for(int j = 0;j < n;j ++) {
            if(i >> j & 1)a += s[n * 2 - 1 - j];
            else b += s[n * 2 - 1 - j];
        }
        ans += mp[pa(a,b)];
    }
    cout<<ans;
}

```