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>
clock_t t=clock();
namespace my_std{
	using namespace std;
	#define pii pair<int,int>
	#define fir first
	#define sec second
	#define MP make_pair
	#define rep(i,x,y) for (int i=(x);i<=(y);i++)
	#define drep(i,x,y) for (int i=(x);i>=(y);i--)
	#define go(x) for (int i=head[x];i;i=edge[i].nxt)
	#define templ template<typename T>
	#define sz 66
	#define mod 1000000007ll
	typedef long long ll;
	typedef double db;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
	templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
	templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
	templ inline void read(T& t)
	{
		t=0;char f=0,ch=getchar();double d=0.1;
		while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
		while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
		if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
		t=(f?-t:t);
	}
	template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
	char __sr[1<<21],__z[20];int __C=-1,__zz=0;
	inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
	inline void print(register int x)
	{
		if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
		while(__z[++__zz]=x%10+48,x/=10);
		while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
	}
	void file()
	{
		#ifdef NTFOrz
		freopen("a.in","r",stdin);
		#endif
	}
	inline void chktime()
	{
		#ifdef NTFOrz
		cout<<(clock()-t)/1000.0<<'\n';
		#endif
	}
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
	ll inv(ll x){return ksm(x,mod-2);}
//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

int n;
ll fac[sz],_fac[sz];

ll dp[sz][sz][sz][sz];

int main()
{
	file();
	read(n);
	fac[0]=_fac[0]=1;rep(i,1,sz-1) _fac[i]=inv(fac[i]=fac[i-1]*i%mod);
	dp[1][1][1][0]=1;
	rep(i,1,n-1) rep(j,1,n) rep(x,0,j) rep(y,0,j-x) if (dp[i][j][x][y]) rep(p,0,x+y) rep(s,0,n-j)
	{
		// (p,x+y-p) \to (x,y), next_sum=s
		int k=p-x;
		if ((s-k+100000)&1) continue;
		ll w=dp[i][j][x][y]*_fac[p]%mod*_fac[x+y-p]%mod;
		if (k<=0)
		{
			k=-k;
			int a=k+(s-k)/2,b=(s-k)/2;
			if ((!a&&!b)||x||y) (dp[i+1][j+s][a][b]+=w)%=mod;
		}
		else
		{
			int a=(s-k)/2,b=k+(s-k)/2;
			if ((!a&&!b)||x||y) (dp[i+1][j+s][a][b]+=w)%=mod;
		}
	}
	ll ans=0;
	rep(x,0,n) rep(y,0,n-x) (ans+=dp[n][n][x][y]*fac[n]%mod*_fac[x]%mod*_fac[y]%mod)%=mod;
	cout<<ans;
	return 0;
}