#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100000 + 10;
void calc(string s,map<pair<ll,ll>,int> &m){
    int n = s.size();
    for(int i=0;i<1<<n;i++){
        ll x = 0, y = 0;
        for(int j=0;j<n;j++){
            if(i>>j&1) x = x * 233 + (s[j]-'a'+1);
            else y = y * 233 + (s[j]-'a'+1);
        }
        m[{x,y}]++;
    }
}
int main(){
    int n; string s; cin>>n>>s;
    string a,b;
    for(int i=0;i<2*n;i++){
        if(i<n) a += s[i];
        else b += s[i];
    }
    reverse(b.begin(),b.end());
    map<pair<ll,ll>,int> m1,m2;
    calc(a,m1);
    calc(b,m2);
    ll ans = 0;
    for(auto i: m1){
        ans += 1ll * i.second * m2[i.first];
    }
    cout<<ans<<endl;
}

### Prompt

Your task is to create 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;
typedef long long ll;
const int N = 100000 + 10;
void calc(string s,map<pair<ll,ll>,int> &m){
    int n = s.size();
    for(int i=0;i<1<<n;i++){
        ll x = 0, y = 0;
        for(int j=0;j<n;j++){
            if(i>>j&1) x = x * 233 + (s[j]-'a'+1);
            else y = y * 233 + (s[j]-'a'+1);
        }
        m[{x,y}]++;
    }
}
int main(){
    int n; string s; cin>>n>>s;
    string a,b;
    for(int i=0;i<2*n;i++){
        if(i<n) a += s[i];
        else b += s[i];
    }
    reverse(b.begin(),b.end());
    map<pair<ll,ll>,int> m1,m2;
    calc(a,m1);
    calc(b,m2);
    ll ans = 0;
    for(auto i: m1){
        ans += 1ll * i.second * m2[i.first];
    }
    cout<<ans<<endl;
}

```