#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

map<pair<string,string>,ll> m1,m2;
int main(){
    int n;
    string s,t;
    cin>>n>>s;
    t=s.substr(n);
    reverse(t.begin(),t.end());
    s=s.substr(0,n);
    rep(i,1<<n){
        string ts1,tt1,ts2,tt2;
        rep(j,n){
            if(i>>j&1){
                ts1+=s[j],tt1+=t[j];
            }
            else{
                ts2+=s[j],tt2+=t[j];
            }
        }
        ++m1[make_pair(ts1,ts2)],++m2[make_pair(tt1,tt2)];
    }
    ll ans=0;
    for(auto it:m1){
        ans+=it.second*m2[it.first];
    }
    cout<<ans<<endl;
    return 0;
}
### Prompt

In Cpp, your task is to solve 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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

map<pair<string,string>,ll> m1,m2;
int main(){
    int n;
    string s,t;
    cin>>n>>s;
    t=s.substr(n);
    reverse(t.begin(),t.end());
    s=s.substr(0,n);
    rep(i,1<<n){
        string ts1,tt1,ts2,tt2;
        rep(j,n){
            if(i>>j&1){
                ts1+=s[j],tt1+=t[j];
            }
            else{
                ts2+=s[j],tt2+=t[j];
            }
        }
        ++m1[make_pair(ts1,ts2)],++m2[make_pair(tt1,tt2)];
    }
    ll ans=0;
    for(auto it:m1){
        ans+=it.second*m2[it.first];
    }
    cout<<ans<<endl;
    return 0;
}
```