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

typedef pair<string,string> P;

int main() {
    int n;
    long long ans = 0;
    string s;
    map<P,long long> cnt;
    cin >> n >> s;

    for (int i = 0;i < 1 << n;++i) {
        string a = "",b = "";
        for (int j = 0;j < n;++j) {
            if (i & (1 << j)) {
                a += s[j];
            } else b = s[j]+b;
        }
        cnt[P(a,b)]++;
    }

    for (int i = 0;i < 1 << n;++i) {
        string a = "",b = "";
        for (int j = 0;j < n;++j) {
            if (i & (1 << j)) {
                a += s[n+j];
            } else b = s[n+j]+b;
        }
        ans += cnt[P(b,a)];
    }

    cout << ans << endl;
    return 0;
}
### Prompt

Create a solution in Cpp for 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 pair<string,string> P;

int main() {
    int n;
    long long ans = 0;
    string s;
    map<P,long long> cnt;
    cin >> n >> s;

    for (int i = 0;i < 1 << n;++i) {
        string a = "",b = "";
        for (int j = 0;j < n;++j) {
            if (i & (1 << j)) {
                a += s[j];
            } else b = s[j]+b;
        }
        cnt[P(a,b)]++;
    }

    for (int i = 0;i < 1 << n;++i) {
        string a = "",b = "";
        for (int j = 0;j < n;++j) {
            if (i & (1 << j)) {
                a += s[n+j];
            } else b = s[n+j]+b;
        }
        ans += cnt[P(b,a)];
    }

    cout << ans << endl;
    return 0;
}
```