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 title "title"
#define ll long long
#define ull unsigned ll
#define fix(x) fixed<<setprecision(x)
#define pii pair<ll,ll>
#define vint vector<ll>
#define pb push_back
using namespace std;
void Freopen(){
	freopen(title".in","r",stdin);
	freopen(title".out","w",stdout);
}
ll read(){
	ll g=0,f=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch&&ch<='9'){g=g*10+ch-'0';ch=getchar();}
	return g*f;
}
const ll N=55;
const ll mod=1e9+7;
ll fac[N],inv[N],f[N][N],n;
ll C(ll n,ll m){
	if(n<m)return 0;
	return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
signed main(){
	n=read();
	fac[0]=1;for(ll i=1;i<=n;i++)fac[i]=fac[i-1]*i%mod;
	inv[1]=1;for(ll i=2;i<=n;i++)inv[i]=(mod-mod/i)*inv[mod%i]%mod;
	inv[0]=1;for(ll i=1;i<=n;i++)inv[i]=inv[i-1]*inv[i]%mod;
	f[1][0]=f[1][1]=n;
	for(ll i=1;i<=n;i++)for(ll j=1;j<=n-i;j++)for(ll k=0;k<=j;k++){
		if((j-k)&1)continue;ll re=(j-k)/2;
		for(ll d=0;d<=j;d++)f[i+j][abs(re-d)]=(f[i+j][abs(re-d)]+f[i][k]*C(n-i,j)%mod*C(j,d)%mod)%mod;
	}return cout<<f[n][0],signed();
}