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<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef vector<int> vi;
typedef pair<int,int> pii;
const int N=50+5;
const db pi=acos(-1.0);
#define lowbit(x) ((x)&(-x))
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u],v=sq[i].to;i;i=sq[i].nxt,v=sq[i].to)
#define fir first
#define sec second
#define mkp make_pair
#define pb push_back
#define maxd 1000000007
#define eps 1e-8
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

inline ll readll()
{
    ll x=0;int f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

int n,f[N][N],c[N][N];

inline int add(int x,int y) {return x+y>=maxd?x+y-maxd:x+y;}
inline int mul(int x,int y) {return 1ll*x*y%maxd;}

int main()
{
    n=read();
    rep(i,0,n)
    {
        c[i][0]=1;
        rep(j,1,i) c[i][j]=add(c[i-1][j],c[i-1][j-1]);
    }
    f[1][0]=f[1][1]=n;
    rep(i,1,n-1) rep(j,0,i)
    {
        rep(k,j,n-i)
        {
            if ((!k) || (k-j)&1) continue;
            int y=(k-j)/2;
            rep(x,0,k)
            {
                f[i+k][abs(x-y)]=add(f[i+k][abs(x-y)],mul(f[i][j],mul(c[n-i][k],c[k][x])));
            }
        }
    }
    printf("%d\n",f[n][0]);
    return 0;
}