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<algorithm>/*{{{*/
#include<cctype>
#include<cassert>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<vector>
using namespace std;
typedef long long lld;
typedef long double lf;
typedef unsigned long long uld;
typedef pair<int,int> pii;
#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define ROF(i,a,b) for(int i=(a);i>=(b);--i)
namespace RA{
    int r(int p){return 1ll*rand()*rand()%p;}
    int r(int L,int R){return r(R-L+1)+L;}
}/*}}}*/
/******************heading******************/
const int N=52,P=1e9+7;

int n;
int C[N][N],f[N][N];
//f[i,j]表示一共i个点，最后一层有j个点有奇数个儿子的方案数

int main(){
    scanf("%d",&n);
    C[0][0]=1;
    FOR(i,1,n){
        C[i][0]=1;
        FOR(j,1,i){
            C[i][j]=(C[i-1][j-1]+C[i-1][j])%P;
            //printf("C[%d,%d]=%d\n",i,j,C[i][j]);
        }
    }
    f[1][0]=f[1][1]=n;
    FOR(i,1,n){
        FOR(j,0,i){
            FOR(k,max(j,1),n-i){//枚举下一层一共有k个点
                if((k-j)&1)continue;//根据状态定义，k-j必须是偶数
                int t=k-j>>1;
                //假设下一层是叶子结点，那么可以理解为有(k-j)/2个点是-1，(k+j)/2个点是1
                FOR(x,0,k){//枚举下一层实际上有x个点是-1
                    //这样就要求|x-(k-j)/2|个点需要从-1改成1（或者从1改成-1）
                    //相当于下一层有|x-(k-j)/2|个奇数儿子结点
                    f[i+k][abs(x-t)]=(f[i+k][abs(x-t)]+1ll*f[i][j]*C[n-i][k]%P*C[k][x])%P;
                    //printf("(%d,%d) <- (%d,%d)\n",i+k,abs(x-t),i,j);
                }
            }
            //printf("f[%d,%d]=%d\n",i,j,f[i][j]);
        }
    }
    printf("%d\n",f[n][0]);
    return 0;
}
