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 <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long ll;

const int N=55,Mo=1e9+7;

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

void inc(int &x,int y) {x=x+y>=Mo?x+y-Mo:x+y;}

int main() {
	scanf("%d",&n);
	fo(i,0,n) {
		C[i][0]=1;
		fo(j,1,i) C[i][j]=(C[i-1][j]+C[i-1][j-1])%Mo;
	}
	f[1][0]=f[1][1]=n;
	fo(i,1,n-1)
		fo(j,0,i)
			fo(k,max(j,1),n-i) {
				if ((k-j)&1) continue;
				int y=(k-j)/2;// 当前和父亲不一样
				fo(x,0,k) {
					// 目标和父亲不一样
					inc(f[i+k][abs(y-x)],(ll)f[i][j]*C[n-i][k]%Mo*C[k][x]%Mo);
				}
			}
	printf("%d\n",f[n][0]);
	return 0;
}