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<bits/stdc++.h>
#define mod 1000000007
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
int f[51][51],C[51][51];
void add(int&a,int b){a=a+b>=mod?a+b-mod:a+b;}
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi();
	C[0][0]=1;
	for(int i=1;i<=n;++i){
		C[i][0]=1;
		for(int j=1;j<=i;++j)C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
	}
	f[1][0]=f[1][1]=1;
	for(int i=1;i<=n;++i)
		for(int j=0;j<=i;++j)
			for(int x=0;i+x<=n;++x)
				for(int y=0;i+x+y<=n;++y)
					if(x+y&&x+y>=j&&!((x+y-j)&1))
						add(f[i+x+y][abs(j+(x+y-j)/2-x)],1ll*f[i][j]*C[i+x+y][x+y]%mod*C[x+y][y]%mod);
	printf("%d\n",f[n][0]);
	return 0;
}
