#include <iostream>
#include <map>
using namespace std;

using ll = long long;

ll MOD = 1000000007;

int N; string S;

map<pair<string, string>, int> memo;

int main(void) {
    cin >> N >> S;
    for (int i = 0; i < (1 << N); ++i) {
        string s, t;
        for (int j = 0; j < N; ++j)
            (((i >> j) & 1) == 0 ? s : t) += string(1, S[j]);
        memo[make_pair(s, t)] ++;
    }
    ll ans = 0;
    for (int i = 0; i < (1 << N); ++i) {
        string s, t;
        for (int j = 2*N-1; j >= N; --j)
            (((i >> (j-N)) & 1) == 0 ? s : t) += string(1, S[j]);
        ans += memo[make_pair(s, t)];
    }
    cout << ans << endl;
    return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 <iostream>
#include <map>
using namespace std;

using ll = long long;

ll MOD = 1000000007;

int N; string S;

map<pair<string, string>, int> memo;

int main(void) {
    cin >> N >> S;
    for (int i = 0; i < (1 << N); ++i) {
        string s, t;
        for (int j = 0; j < N; ++j)
            (((i >> j) & 1) == 0 ? s : t) += string(1, S[j]);
        memo[make_pair(s, t)] ++;
    }
    ll ans = 0;
    for (int i = 0; i < (1 << N); ++i) {
        string s, t;
        for (int j = 2*N-1; j >= N; --j)
            (((i >> (j-N)) & 1) == 0 ? s : t) += string(1, S[j]);
        ans += memo[make_pair(s, t)];
    }
    cout << ans << endl;
    return 0;
}

```