Problem:
Score : 1600 points

Problem Statement

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.

Sample Input 1

3

Sample Output 1

12

There are 3 checker pieces, positioned at 10^{100}, 10^{200}, 10^{300} respectively. Call them A, B, C respectively.

Here are two (of the 12) possible sequence of moves :

Let A jump over B in the first second, and let A jump over C in the second second. The final position of A is 2 \times 10^{300} - 2 \times 10^{200} + 10^{100}.

Let C jump over A in the first second, and let B jump over C in the second second. The final position of B is -2 \times 10^{300} - 10^{200} + 4 \times 10^{100}.

There are a total of 3 \times 2 \times 2 = 12 possible sequence of moves, and the final piece are in different positions in all of them.

Sample Input 2

4

Sample Output 2

84

Sample Input 3

22

Sample Output 3

487772376

Remember to output your answer modulo 10^{9} + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<P,int> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

ll Pow(ll n,ll p){
	ll r=1;
	for(;p>0;p>>=1){
		if(p&1) r=(r*n)%mod;
		n=(n*n)%mod;
	}
	return r;
}

const int M=55;
ll F[M];

void Init(){
	F[0]=1;
	for(int i=1;i<M;i++) F[i]=F[i-1]*i%mod;
}

ll Div(ll n,ll m){
	return n*Pow(m,mod-2)%mod;
}

ll n,dp[M][M][M];

int main(){
	cin>>n;
	Init();
	dp[0][1][0]=F[n];
	dp[0][2][1]=mod-F[n];
	for(int i=0;i<n;i++){
		for(int j=0;j<M;j++) for(int k=1;j+k<M;k++) (dp[i][j+k][k]+=dp[i][j+k-1][k-1])%=mod;
		for(int j=1;j<M;j++) for(int k=1;j+k<M;k++) (dp[i][k][j+k]+=dp[i][k-1][j+k-1])%=mod;
		for(int j=0;j<M;j++) for(int k=0;k<M;k++) if(i+j+k<M){
			for(int l=0;l<=j;l++) (dp[i+j+k][l][0]+=Div(dp[i][j][k],F[j-l]*F[k+l]%mod))%=mod;
			for(int l=1;l<=k;l++) (dp[i+j+k][0][l]+=Div(dp[i][j][k],F[j+l]*F[k-l]%mod))%=mod;
		}
	}
	cout<<dp[n][0][0]<<endl;
}