#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
int n;
char s[40];
map<pair<string, string>, ll> mp;

int main() {
    scanf("%d %s", &n, s);
    for (int i = 0; i < 1 << n; i++) {
        string a, b;
        for (int j = 0; j < n; j++) {
            if (i >> j & 1) a.push_back(s[j]);
            else b.push_back(s[j]);
        }
        mp[{a, b}]++;
    }
    ll ans = 0;
    for (int i = 0; i < 1 << n; i++) {
        string a, b;
        for (int j = 0; j < n; j++) {
            if (i >> j & 1) a.push_back(s[2 * n - 1 - j]);
            else b.push_back(s[2 * n - 1 - j]);
        }
        ans += mp[{a, b}];
    }
    printf("%lld\n", ans);
    return 0;
}
### 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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
int n;
char s[40];
map<pair<string, string>, ll> mp;

int main() {
    scanf("%d %s", &n, s);
    for (int i = 0; i < 1 << n; i++) {
        string a, b;
        for (int j = 0; j < n; j++) {
            if (i >> j & 1) a.push_back(s[j]);
            else b.push_back(s[j]);
        }
        mp[{a, b}]++;
    }
    ll ans = 0;
    for (int i = 0; i < 1 << n; i++) {
        string a, b;
        for (int j = 0; j < n; j++) {
            if (i >> j & 1) a.push_back(s[2 * n - 1 - j]);
            else b.push_back(s[2 * n - 1 - j]);
        }
        ans += mp[{a, b}];
    }
    printf("%lld\n", ans);
    return 0;
}
```