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
//by yjz
#include<bits/stdc++.h>
using namespace std;
#define FF first
#define SS second
#define PB push_back
#define MP make_pair
#define foreach(it,s) for(__typeof((s).begin()) it=(s).begin();it!=(s).end();it++)
#ifndef LOCAL
#define cerr if(0)cout
#endif
typedef long long ll;
const int mod=1e9+7;
ll qpow(ll x, ll k) {return k==0? 1: 1ll*qpow(1ll*x*x%mod,k>>1)*(k&1?x:1)%mod;}
int n;
ll fac[55], ifac[55];
//tot c+ c-
int dp[55][55][55];
bool check(int a, int b, int c, int d)
{
	if (c<d) return d-c<=b;
	else return c-d<=a;
}
void upd(int &x, ll v)
{
	x = (x+v)%mod;
}
int main()
{
	fac[0] = 1;
	for (int i=1; i<=50; i++) fac[i] = fac[i-1]*i%mod;
	for (int i=0; i<=50; i++) ifac[i] = qpow(fac[i], mod-2);
	cin>>n;
	dp[1][0][0] = 1;
	dp[1][1][0] = 1;
	for (int i=1; i<n; i++)
	{
		for (int cp=0; cp<=i; cp++)
		{
			for (int cn=0; cn+cp<=i; cn++)
			{
				if (cp>0&&cn>0) continue;
				int &cur = dp[i][cp][cn];
				for (int ex=0; cp+cn+ex+ex+i<=n; ex++)
				{
					if (cp==0&&cn==0&&ex==0) continue;
					for (int x=0; x<=cp+ex; x++)
					{
						for (int y=0; y<=cn+ex; y++)
						{
							upd(dp[i+cp+cn+ex+ex][x][y], 1ll*cur*ifac[cp+ex-x+y]%mod*ifac[cn+ex-y+x]%mod);
						}
					}
				}
			}
		}
	}
	ll ans = dp[n][0][0];
	cout<<1ll*ans*fac[n]%mod<<endl;
	return 0;
}
