#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

int n;
string s;

map<pair<string, string>, ll> mp;

string a, b;
void rec1(int i = 0) {
  if (i == n) {
    mp[{a, b}]++;
    return;
  }
  a.push_back(s[i]);
  rec1(i + 1);
  a.pop_back();
  b.push_back(s[i]);
  rec1(i + 1);
  b.pop_back();
}

ll res = 0;
void rec2(int i = 2 * n) {
  if (i == n) {
    res += mp[{a, b}];
    return;
  }
  a.push_back(s[i - 1]);
  rec2(i - 1);
  a.pop_back();
  b.push_back(s[i - 1]);
  rec2(i - 1);
  b.pop_back();
}

signed main() {
#ifdef LOCAL
  assert(freopen("c.in", "r", stdin));
#endif
  cin >> n >> s;
  rec1();
  rec2();
  cout << res << '\n';
}

### Prompt

Generate 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;
using ll = long long;
using ld = long double;

int n;
string s;

map<pair<string, string>, ll> mp;

string a, b;
void rec1(int i = 0) {
  if (i == n) {
    mp[{a, b}]++;
    return;
  }
  a.push_back(s[i]);
  rec1(i + 1);
  a.pop_back();
  b.push_back(s[i]);
  rec1(i + 1);
  b.pop_back();
}

ll res = 0;
void rec2(int i = 2 * n) {
  if (i == n) {
    res += mp[{a, b}];
    return;
  }
  a.push_back(s[i - 1]);
  rec2(i - 1);
  a.pop_back();
  b.push_back(s[i - 1]);
  rec2(i - 1);
  b.pop_back();
}

signed main() {
#ifdef LOCAL
  assert(freopen("c.in", "r", stdin));
#endif
  cin >> n >> s;
  rec1();
  rec2();
  cout << res << '\n';
}

```