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
// This amazing code is by Eric Sunli Chen.
#include<bits/stdc++.h>
using namespace std;
template<typename T> bool get_int(T &x)
{
	char t=getchar();
	bool neg=false;
	x=0;
	for(; (t>'9'||t<'0')&&t!='-'&&t!=EOF; t=getchar());
	if(t=='-')neg=true,t=getchar();if(t==EOF)return false;
	for(; t<='9'&&t>='0'; t=getchar())x=x*10+t-'0';
	if(neg)x=-x;return true;
}
template<typename T> void print_int(T x)
{
	if(x<0)putchar('-'),x=-x;
	short a[20]= {},sz=0;
	while(x>0)a[sz++]=x%10,x/=10;
	if(sz==0)putchar('0');
	for(int i=sz-1; i>=0; i--)putchar('0'+a[i]);
}
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define get1(a) get_int(a)
#define get2(a,b) (get1(a)&&get1(b))
#define get3(a,b,c) (get1(a)&&get2(b,c))
#define printendl(a) print_int(a),puts("")
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int,int> pii;
typedef vector<int> vi;
const int inf=0x3f3f3f3f;
const LL Linf=1ll<<61;
const double pi=acos(-1.0);

const int mod=1e9+7;
const int maxn=55;
inline void add(int&x,int y){x+=y;if(x>=mod)x-=mod;}


int n,dp[maxn][maxn],c[maxn][maxn];
int main()
{
	for(int i=0;i<maxn;i++)
	{
		c[i][0]=1;
		for(int j=1;j<=i;j++)
		{
			c[i][j]=c[i-1][j]+c[i-1][j-1];
			if(c[i][j]>=mod)c[i][j]-=mod;
		}
	}
	scanf("%d",&n);
	dp[1][0]=n;dp[1][1]=n;
	for(int i=1;i<=n;i++)for(int j=0;j<=i+1;j++)if(dp[i][j])
	{
		for(int k1=0;k1<=n-i;k1++)for(int k2=0;k2<=n-i;k2++)if(i+k1+k2<=n&&k1+k2>=max(j,1)&&(k1+k2+j)%2==0)
			add(dp[i+k1+k2][abs(j+k1-k2)/2],(LL)dp[i][j]*c[n-i][k1+k2]%mod*c[k1+k2][k1]%mod);
	}
	printf("%d\n",dp[n][0]);
	return 0;
}