Problem:
Score : 1600 points

Problem Statement

Let X = 10^{100}. Inaba has N checker pieces on the number line, where the i-th checker piece is at coordinate X^{i} for all 1 \leq i \leq N.

Every second, Inaba chooses two checker pieces, A and B, and move A to the symmetric point of its current position with respect to B. After that, B is removed. (It is possible that A and B occupy the same position, and it is also possible for A to occupy the same position as another checker piece after the move).

After N - 1 seconds, only one checker piece will remain. Find the number of distinct possible positions of that checker piece, modulo 10^{9} + 7.

Constraints

1 \leq N \leq 50

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of distinct possible positions of the final checker piece, modulo 10^{9} + 7.

Sample Input 1

3

Sample Output 1

12

There are 3 checker pieces, positioned at 10^{100}, 10^{200}, 10^{300} respectively. Call them A, B, C respectively.

Here are two (of the 12) possible sequence of moves :

Let A jump over B in the first second, and let A jump over C in the second second. The final position of A is 2 \times 10^{300} - 2 \times 10^{200} + 10^{100}.

Let C jump over A in the first second, and let B jump over C in the second second. The final position of B is -2 \times 10^{300} - 10^{200} + 4 \times 10^{100}.

There are a total of 3 \times 2 \times 2 = 12 possible sequence of moves, and the final piece are in different positions in all of them.

Sample Input 2

4

Sample Output 2

84

Sample Input 3

22

Sample Output 3

487772376

Remember to output your answer modulo 10^{9} + 7.

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

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<cstdlib>
#define cmax(a,b) (a<(b)?a=(b),1:0)
#define cmin(a,b) (a>(b)?a=(b),1:0)
#define dmin(a,b) ((a)<(b)?(a):(b))
#define dmax(a,b) ((a)>(b)?(a):(b))
#define CL fclose(stdin),fclose(stdout)
namespace io
{
	int F()
	{
		int n=0,F=1;
		char ch;
		while((ch=getchar())!='-'&&(ch<'0'||ch>'9'));
		ch=='-'?F=0:n=ch-'0';
		while((ch=getchar())>='0'&&ch<='9')n=(n<<1)+(n<<3)+ch-'0';
		return F?n:-n;
	}
	long long G()
	{
		long long n=0,F=1;
		char ch;
		while((ch=getchar())!='-'&&(ch<'0'||ch>'9'));
		ch=='-'?F=0:n=ch-'0';
		while((ch=getchar())>='0'&&ch<='9')n=(n<<1)+(n<<3)+ch-'0';
		return F?n:-n;
	}
}
int c[555][555];
int f[555][555][2];
int main()
{
	//freopen("rabbit.in","r",stdin);
	//freopen("rabbit.out","w",stdout);
	int n=io::F(),M=1000000007;
	for(register int i=0;i<=n;++i)c[i][0]=c[i][i]=1;
	for(register int i=2;i<=n;++i)
		for(register int j=1;j<i;++j)
			c[i][j]=(c[i-1][j-1]+c[i-1][j])%M;
	f[1][0][0]=c[n][1];
	f[1][1][0]=c[n][1];
	for(register int i=1;i<n;++i)
		for(register int j=0;j<=i;++j)
			for(register int k=0;k<=1;++k)
				if(f[i][j][k])
				{
					long long F=f[i][j][k];
					for(register int x=0;x+i<=n;++x)
					{
						register int y=x==0?1:0;
						if((x+y^j)&1)++y;
						for(;x+y+i<=n;y+=2)
						{
							int cnt=x+y;
							int ck=j+(cnt-j)/2,ct=cnt-ck;
							if(ck>=x)f[i+cnt][ck-x][k]=(f[i+cnt][ck-x][k]+F*c[n-i][x]%M*c[n-i-x][y])%M;
								else f[i+cnt][x-ck][!k]=(f[i+cnt][x-ck][!k]+F*c[n-i][x]%M*c[n-i-x][y])%M;
						}
					}
				}
	printf("%d\n",f[n][0][0]);
	return 0;
}