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<cmath>
#include<math.h>
#include<ctype.h>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<complex>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cwchar>
#include<cwctype>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
const long long Mod=1e9+7;
int n;
long long f[55][110],c[55][55];
int main()
{
	scanf("%d",&n);
	for (int i=0;i<=n;i++)
	{
		c[i][0]=1;
		for (int j=1;j<=i;j++)
		{
			c[i][j]=(c[i-1][j-1]+c[i-1][j])%Mod;
		}
	}
	f[1][n]=f[1][n-1]=1;
	for (int i=1;i<=n;i++)
	{
		for (int j=-i;j<=i;j++)
		{
			for (int a=0;a<n;a++)
			{
				for (int b=0;b<n;b++)
				{
					if (a+b==0 || a+b+i>n || (a+b+j)&1 || abs(j)>a+b)
					{
						continue;
					}
					f[i+a+b][(j+a-b)/2+n]=(f[i+a+b][(j+a-b)/2+n]+c[a+b+i][a+b]*c[a+b][a]%Mod*f[i][j+n]%Mod)%Mod;
				}
			}
		}
	}
	printf("%lld\n",f[n][n]);
    return 0;
}
//曾经沧海难为水，除却巫山不是云。
//——元稹《离思》