#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
int main()
{
	int n;
	string s;
	cin>>n>>s;
	ll ans=0;
	for(int k=0; k<(1<<n); k++){
		string t1, t2;
		for(int i=0; i<n; i++){
			if(k&(1<<i)){
				t1+=s[i];
			}else{
				t2+=s[i];
            		}
		}
		ll dp[19][19]={};
		int m=t1.size();
		dp[0][0]=1;
		for(int i=1; i<=n; i++){
			for(int j=0; j<=m; j++){
				if(j>=1 && s[n+i-1]==t1[m-j]){
					dp[i][j]+=dp[i-1][j-1];
				}
				if(i-j-1>=0 && s[n+i-1]==t2[n-m-i+j]){
					dp[i][j]+=dp[i-1][j];
				}
			}
		}
		ans+=dp[n][m];
	}
	cout<<ans<<endl;
    return 0;
}
### Prompt

Please provide a CPP coded solution 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 <bits/stdc++.h>
using namespace std;
typedef long long int ll;
int main()
{
	int n;
	string s;
	cin>>n>>s;
	ll ans=0;
	for(int k=0; k<(1<<n); k++){
		string t1, t2;
		for(int i=0; i<n; i++){
			if(k&(1<<i)){
				t1+=s[i];
			}else{
				t2+=s[i];
            		}
		}
		ll dp[19][19]={};
		int m=t1.size();
		dp[0][0]=1;
		for(int i=1; i<=n; i++){
			for(int j=0; j<=m; j++){
				if(j>=1 && s[n+i-1]==t1[m-j]){
					dp[i][j]+=dp[i-1][j-1];
				}
				if(i-j-1>=0 && s[n+i-1]==t2[n-m-i+j]){
					dp[i][j]+=dp[i-1][j];
				}
			}
		}
		ans+=dp[n][m];
	}
	cout<<ans<<endl;
    return 0;
}
```