Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

4
cabaacba

Sample Output 1

4

There are four ways to paint the string, as follows:

cabaacba

cabaacba

cabaacba

cabaacba

Sample Input 2

11
mippiisssisssiipsspiim

Sample Output 2

504

Sample Input 3

4
abcdefgh

Sample Output 3

0

Sample Input 4

18
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Sample Output 4

9075135300

The answer may not be representable as a 32-bit integer.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#define _SCL_SECURE_NO_WARNINGS
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <utility>
#include <algorithm>
#include <functional>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <limits>
#include <numeric>
#include <valarray>
#include <fstream>

using namespace std;
typedef unsigned int uint;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, LL> PP;
#define REP(i, a, n) for(LL i = (a), i##_max = (n); i < i##_max; ++i)
#define REM(i, a, n) for(LL i = (LL)(n) - 1, i##min = (a); i >= i##min; --i)
#define ALL(arr) (arr).begin(), (arr).end()
#define FLOAT fixed << setprecision(16)
#define SPEEDUP {cin.tie(NULL); ios::sync_with_stdio(false);}
const int INF = 0x3FFFFFFF;
const LL INFLL = 0x3FFFFFFF3FFFFFFF;
const double INFD = 1.0e+308;
const string INFSTR = "\x7f";
const double EPS = 1.0e-9;

void YesNo(bool b) { cout << (b ? "Yes" : "No") << endl; }
void YESNO(bool b) { cout << (b ? "YES" : "NO") << endl; }
template <class T, class U>
istream& operator>>(istream& ist, pair<T, U>& right) { return ist >> right.first >> right.second; }
template <class T, class U>
ostream& operator<<(ostream& ost, const pair<T, U>& right) { return ost << right.first << ' ' << right.second; }
template <class T, class TCompatible, size_t N>
void Fill(T(&dest)[N], const TCompatible& val) { fill(dest, dest + N, val); }
template <class T, class TCompatible, size_t M, size_t N>
void Fill(T(&dest)[M][N], const TCompatible& val) { for (int i = 0; i < M; ++i) Fill(dest[i], val); }
template<class T>
T Compare(T left, T right) { return left > right ? 1 : (left < right ? -1 : 0); }
istream& Ignore(istream& ist) { string s; ist >> s; return ist; }
bool Inside(int i, int j, int h, int w) { return i >= 0 && i < h && j >= 0 && j < w; }
template <class T>
T Next() { T buf; cin >> buf; return buf; }

#ifdef ONLY_MY_ENVIR
#include "IntMod.h"
#include "Union_Find.h"
#include "Graph.h"
#include "Range.h"
#include "Global.h"
#include "Flow_Solver.h"
#include "Tree.h"
#include "Suffix_Array.h"
#include "Geometry.h"
#include "Matrix.h"
#include "Segment_Tree.h"
#include "Rational.h"
#include "Position.h"
#include "Factorization.h"
#endif

#ifdef __GNUC__
typedef __int128 LLL;
istream& operator>> (istream& ist, __int128& val) { LL tmp;  ist >> tmp; val = tmp; return ist; }
ostream& operator<< (ostream& ost, __int128 val) { LL tmp = val; ost << tmp; return ost; }
#endif

#if 1234567891
#include <array>
#include <random>
#include <unordered_set>
#include <unordered_map>
template<typename T>
using PriorityQ = priority_queue<T, vector<T>, greater<T> >;	// コスト小を優先
template <class T>
auto Is(const T& value) { return [value](const auto& comparand) -> bool { return comparand == value; }; }
#endif

using PS = pair<string, string>;
int N;
string s;
map<PS, LL> mp;

int main() {	
	cin >> N;
	cin >> s;
	REP(i, 0, 1 << N) {
		string p;
		string q;
		REP(k, 0, N) {
			if ((i >> k) & 1) {
				p += s[k];
			} else {
				q += s[k];
			}
		}
		++mp[PS(p, q)];
	}

	reverse(s.begin(), s.end());
	LL ans = 0;
	REP(i, 0, 1 << N) {
		string p;
		string q;
		REP(k, 0, N) {
			if ((i >> k) & 1) {
				p += s[k];
			} else {
				q += s[k];
			}
		}
		PS ps(p, q);
		if (mp.count(ps)) {
			ans += mp[ps];
		}
	}
	cout << ans << endl;
	return 0;
}