#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 111;

char s[N];
int main() {
    map<pair<string,string>,int> ct;
    int n;
    scanf("%d%s", &n, s);
    for (int m = 0; m < 1 << n; m++) {
        string r, b;
        for (int i = 0; i < n; i++) {
            (m & (1 << i) ? r : b).push_back(s[i]);
        }
        ct[{r, b}]++;
    }
    ll ans = 0;
    for (int m = 0; m < 1 << n; m++) {
        string r, b;
        for (int i = 0; i < n; i++) {
            (m & (1 << i) ? r : b).push_back(s[i + n]);
        }
        reverse(r.begin(), r.end());
        reverse(b.begin(), b.end());
        ans += ct[{b, r}];
    }
    printf("%lld\n", ans);
}

### 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 = 111;

char s[N];
int main() {
    map<pair<string,string>,int> ct;
    int n;
    scanf("%d%s", &n, s);
    for (int m = 0; m < 1 << n; m++) {
        string r, b;
        for (int i = 0; i < n; i++) {
            (m & (1 << i) ? r : b).push_back(s[i]);
        }
        ct[{r, b}]++;
    }
    ll ans = 0;
    for (int m = 0; m < 1 << n; m++) {
        string r, b;
        for (int i = 0; i < n; i++) {
            (m & (1 << i) ? r : b).push_back(s[i + n]);
        }
        reverse(r.begin(), r.end());
        reverse(b.begin(), b.end());
        ans += ct[{b, r}];
    }
    printf("%lld\n", ans);
}

```