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 pb push_back
#define mp make_pair
#define fi first
#define se second
#define LL long long
#define FOR(i,a,b) for (int i=a;i<=b;++i)
#define FORD(i,a,b) for (int i=a;i>=b;--i)
using namespace std;
typedef pair<int,int> pa;
void getint(int &v){
    char ch,fu=0;
    for(ch='*'; (ch<'0'||ch>'9')&&ch!='-'; ch=getchar());
    if(ch=='-') fu=1, ch=getchar();
    for(v=0; ch>='0'&&ch<='9'; ch=getchar()) v=v*10+ch-'0';
    if(fu) v=-v;
}
const int MO=1e9+7;
int n;
LL f[111][111][111],C[111][111];
int main(){
	cin>>n;
	C[0][0]=1;
	FOR(i,1,n){
		C[i][0]=1;
		FOR(j,1,i) C[i][j]=(C[i-1][j]+C[i-1][j-1])%MO;
	}
	f[0][0][0]=1;
	FOR(i,1,n)
		FOR(a,0,i)
			FOR(b,0,i-a){
				if (a==0 && b==0) continue;
				FOR(x,0,a+b){
					LL t=0;
					int d=x-a;
					FOR(A,0,i-a-b){
						int B=A+d;
						if (B<0) continue;
						(t+=f[i-a-b][A][B])%=MO;
					}
					(f[i][a][b]+=t*C[i][x]%MO*C[i-x][a+b-x])%=MO;
				}
			}
	cout<<(f[n][0][1]%MO+MO)%MO<<endl;
	return 0;
}