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 sqr(x) ((x)*(x))
#define ll long long
#define ull unsigned long long
#define ui unsigned int
#define ld long double
#define vi vector<int>
#define all(a) a.begin(),a.end()
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > > 
#define pii pair<int,int> 
#define pb push_back
#define mp make_pair
#define debuge cerr<<"isok"<<endl
#define debug(x) cerr<<#x<<"="<<x<<endl
#define dprintf(...) fprintf(stderr,__VA_ARGS__)
#define SS second
#define FF first
#define ls (k<<1)
#define rs (k<<1|1)
#define clr(a,x) memset(a,x,sizeof(a))
#define cpy(a,x) memcpy(a,x,sizeof(a))
#define file(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
#define SZ(x) ((int)x.size())
using namespace std;
template<class T> inline void gmin(T &x,const T &y){x=x>y?y:x;}
template<class T> inline void gmax(T &x,const T &y){x=x<y?y:x;}
template<class T> inline bool Gmin(T &x,const T &y){return x>y?x=y,1:0;}
template<class T> inline bool Gmax(T &x,const T &y){return x<y?x=y,1:0;}
const int BufferSize=1<<16;
char buffer[BufferSize],*Bufferhead,*Buffertail;
bool Terminal;
inline char Getchar(){
	if(Bufferhead==Buffertail){
		int l=fread(buffer,1,BufferSize,stdin);
		if(!l){Terminal=1;return 0;}
		Buffertail=(Bufferhead=buffer)+l;
	}
	return *Bufferhead++;
}
template<class T>inline bool read(T &x){
	x=0;char c=Getchar(),rev=0;
	while(c<'0'||c>'9'){rev|=c=='-';c=Getchar();if(Terminal)return 0;}
	while(c>='0'&&c<='9') x=x*10+c-'0',c=Getchar();
	if(c=='.'){
		c=Getchar();double t=0.1;
		while(c>='0'&&c<='9') x=x+(c-'0')*t,c=Getchar(),t=t/10;
	}
	x=rev?-x:x;
	return 1;
}
template<class T1,class T2> inline bool read(T1 &x,T2 &y){return read(x)&read(y);}
template<class T1,class T2,class T3> inline bool read(T1 &x,T2 &y,T3 &z){return read(x)&read(y)&read(z);}
template<class T1,class T2,class T3,class T4> inline bool read(T1 &x,T2 &y,T3 &z,T4 &w){return read(x)&read(y)&read(z)&read(w);}
inline bool reads(char *x){
	char c=Getchar();
	while(c<33||c>126){c=Getchar();if(Terminal)return 0;}
	while(c>=33&&c<=126) (*x++)=c,c=Getchar();
	*x=0;return 1;
}
template<class T>inline void print(T x,const char c='\n'){
	if(!x){putchar('0');putchar(c);return;}
	if(x<0) putchar('-'),x=-x;
	int m=0,a[20];
	while(x) a[m++]=x%10,x/=10;
	while(m--) putchar(a[m]+'0');
	putchar(c);
}
//--------------------------------head---------------------------------------------

const int inf=0x3f3f3f3f;
const int N=55,M=100005,mod=1e9+7;
template<class T,class S> inline void ch(T &x,const S y){x=(x+y)%mod;}
inline int exp(int x,int y,const int mod=::mod){
	int ans=1;
	while(y){
		if(y&1) ans=(ll)ans*x%mod;
		x=(ll)x*x%mod;y>>=1;
	}return ans;
}

int n,ans,f[2][N][N][N],fac[N],ifac[N];

inline int C(int x,int y){
	return x<y?0:(ll)fac[x]*ifac[y]%mod*ifac[x-y]%mod;
}

int main(){
#ifdef rqgao2014
	assert(freopen("input.txt","r",stdin));
#endif
	scanf("%d",&n);fac[0]=1;
	for(int i=1;i<=n;i++) fac[i]=(ll)fac[i-1]*i%mod;
	ifac[n]=exp(fac[n],mod-2);
	for(int i=n;i;i--) ifac[i-1]=(ll)ifac[i]*i%mod;
	f[0][0][1][0]=1;
	for(int i=1;i<=n;i++){
		clr(f[i&1],0);
		for(int j=i;j<=n;j++)
			for(int a=0;a<=n-j;a++)
				for(int b=0;b<=n-j-a;b++){
				
					for(int ap=max(0,a-b);ap<=j;ap++)
						for(int bp=max(0,b-a);bp<=j-ap;bp++)
							ch(f[i&1][j][a][b],(ll)f[i-1&1][j-ap-bp][ap][bp]*fac[j]%mod*ifac[j-ap-bp]%mod*ifac[ap-(a-b)]%mod*ifac[bp+(a-b)]);
//					if(f[i&1][j][a][b]) dprintf("%d %d %d %d %d\n",i,j,a,b,f[i&1][j][a][b]);
				}
		ch(ans,f[i&1][n][0][0]);
		for(int j=i;j<=n;j++)
			f[i&1][j][0][0]=0;
	}
	printf("%d\n",ans);
	return 0;
}