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.

Examples

Input

3


Output

12


Input

4


Output

84


Input

22


Output

487772376
#include<cstdio>
#include<algorithm>
using namespace std;
#define MN 50
#define MOD 1000000007
int c[MN+5][MN+5],f[MN+5][2][MN+5];
inline void rw(int&a,int b){if((a+=b)>=MOD)a-=MOD;}
int main()
{
	int n,i,j,k,l;
	scanf("%d",&n);
	for(i=0;i<=n;++i)for(c[i][0]=j=1;j<=i;++j)c[i][j]=(c[i-1][j-1]+c[i-1][j])%MOD;
	f[1][0][0]=f[1][1][1]=n;
	for(i=1;i<n;++i)for(j=0;j<=i;++j)
		for(k=0;i+k<=n;++k)for(l=(j+k)&1;i+k+l<=n;l+=2)if(k+l&&k+l>=j)
		{
			if((k+l+j)/2<k)rw(f[i+k+l][1][k-(k+l+j)/2],1LL*f[i][0][j]*c[n-i][k]%MOD*c[n-i-k][l]%MOD);
			else rw(f[i+k+l][0][(k+l+j)/2-k],1LL*f[i][0][j]*c[n-i][k]%MOD*c[n-i-k][l]%MOD);
			if((k+l-j)/2<k)rw(f[i+k+l][1][k-(k+l-j)/2],1LL*f[i][1][j]*c[n-i][k]%MOD*c[n-i-k][l]%MOD);
			else rw(f[i+k+l][0][(k+l-j)/2-k],1LL*f[i][1][j]*c[n-i][k]%MOD*c[n-i-k][l]%MOD);
		}
	printf("%d",f[n][0][0]);
}