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 <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>

#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second

using namespace std;

template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }

typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;

const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+7;
const int MAXN=55;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{
	int f=1,x=0; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }
	while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }
	return x*f;
}
int f[MAXN][MAXN],c[MAXN][MAXN];
int upd(int x,int y){ return x+y>=mods?x+y-mods:x+y; }
int main()
{
	int n=read();
	f[1][0]=f[1][1]=1;
	for (int i=0;i<=n;i++) c[i][0]=1;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=i;j++) c[i][j]=upd(c[i-1][j-1],c[i-1][j]);
	for (int i=1;i<n;i++)
		for (int j=0;j<=i;j++) 
			if (f[i][j]) 
			for (int x=0;x+i<=n;++x)
				for (int y=0;x+y+i<=n;++y) 
					if ((x+y)&&(x+y>=j)&&(!((x+y-j)&1))) 
						f[i+x+y][abs((x+y-j)/2+j-x)]=upd(f[i+x+y][abs((x+y-j)/2+j-x)],1ll*f[i][j]*c[i+x+y][x+y]%mods*c[x+y][x]%mods);
	printf("%d\n",f[n][0]);
	return 0;
}
