#include <bits/stdc++.h>
#define FOR(i,k,n)  for(int i = (k);i < (n);++i)
#define REP(i,n)    FOR(i,0,n)
#define ALL(x)      begin(x),end(x)

using namespace std;
using vecint = vector<int>;
using ll = int64_t;

int main()
{
  int n;
  cin>>n;
  string s;
  cin>>s;
  map<pair<string, string>, int> m;
  REP(i,1<<n) {
    string l, r;
    REP(j,n) {
      if ((i>>j) & 1) {
        l += s[j];
      } else {
        r = s[j] + r;
      }
    }
    ++m[make_pair(l, r)];
  }
  ll cnt = 0;
  REP(i,1<<n) {
    string l, r;
    REP(j,n) {
      if ((i>>j) & 1) {
        l += s[j+n];
      } else {
        r = s[j+n] + r;
      }
    }
    if (m.count(make_pair(r, l))) {
      cnt += m[make_pair(r, l)];
    }
  }
  cout << cnt << endl;
  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>
#define FOR(i,k,n)  for(int i = (k);i < (n);++i)
#define REP(i,n)    FOR(i,0,n)
#define ALL(x)      begin(x),end(x)

using namespace std;
using vecint = vector<int>;
using ll = int64_t;

int main()
{
  int n;
  cin>>n;
  string s;
  cin>>s;
  map<pair<string, string>, int> m;
  REP(i,1<<n) {
    string l, r;
    REP(j,n) {
      if ((i>>j) & 1) {
        l += s[j];
      } else {
        r = s[j] + r;
      }
    }
    ++m[make_pair(l, r)];
  }
  ll cnt = 0;
  REP(i,1<<n) {
    string l, r;
    REP(j,n) {
      if ((i>>j) & 1) {
        l += s[j+n];
      } else {
        r = s[j+n] + r;
      }
    }
    if (m.count(make_pair(r, l))) {
      cnt += m[make_pair(r, l)];
    }
  }
  cout << cnt << endl;
  return 0;
}

```