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: Accepted)

#include<bits/stdc++.h>
const int N=55,mo=1e9+7;
int n,i,j,k,l,o,x,d,f[N][N][N],fac[N],inv[N],vf[N];
int main(){
	for(i=*fac=*vf=1;i<N;++i)fac[i]=1ll*fac[i-1]*i%mo,
		inv[i]=i==1?1:1ll*(mo-mo/i)*inv[mo%i]%mo,vf[i]=1ll*vf[i-1]*inv[i]%mo;
	scanf("%d",&n);for(i=0;i<=n;++i)for(j=0;i+j<=n;++j)f[i+j][i][j]=1ll*vf[i]*vf[j]%mo;
	for(i=1;i<n;++i)for(j=0;j<=i;++j)for(k=0;j+k<=i;++k)if(x=f[i][j][k])
		for(d=j-k,l=d>0?0:-d;i+l<=n;++l)for(o=d>0?d:0;i+l+o<=n;++o)if(l|o)
			f[i+l+o][l+d][o-d]=(f[i+l+o][l+d][o-d]+1ll*vf[l]*vf[o]%mo*x)%mo;
	printf("%lld\n",1ll*fac[n]*f[n][1][0]%mo);
	return 0;
}