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)

// #pragma GCC target("avx")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
// #define BEGIN_STACK_EXTEND(size) void * stack_extend_memory_ = malloc(size);void * stack_extend_origin_memory_;char * stack_extend_dummy_memory_ = (char*)alloca((1+(int)(((long long)stack_extend_memory_)&127))*16);*stack_extend_dummy_memory_ = 0;asm volatile("mov %%rsp, %%rbx\nmov %%rax, %%rsp":"=b"(stack_extend_origin_memory_):"a"((char*)stack_extend_memory_+(size)-1024));
// #define END_STACK_EXTEND asm volatile("mov %%rax, %%rsp"::"a"(stack_extend_origin_memory_));free(stack_extend_memory_);

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<unordered_map>
#include<complex>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const long double EPS=1e-9;
const long double INF=1e+10;
const long double PI=acos(-1.0);
const int C_SIZE = 3100000;
const int UF_SIZE = 3100000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
	int UF[UF_SIZE];
	void init_UF(int n){
		for(int i=0;i<n;i++)UF[i]=-1;
	}
	int FIND(int a){
		if(UF[a]<0)return a;
		return UF[a]=FIND(UF[a]);
	}
	void UNION(int a,int b){
		a=FIND(a);b=FIND(b);if(a==b)return;
		// if(UF[a]>UF[b])swap(a,b);
		UF[a]+=UF[b];UF[b]=a;
	}
}
// ここから編集しろ
long long dp[52][52][52][52];
int main(){
	int a;scanf("%d",&a);
	init_C(110);
	dp[1][1][1][0]=1;
	for(int i=1;i<a;i++){
		for(int j=0;j<=a;j++){
			for(int k=0;k<=a;k++){
				for(int l=0;l<=a;l++){
					if(!dp[i][j][k][l])continue;
					// printf("%d %d %d %d: %lld\n",i,j,k,l,dp[i][j][k][l]);
					for(int m=0;j+m<=a;m++){
						for(int n=0;j+m+n<=a;n++){
							int tk=k;
							int tl=l;
							if((m+n)&&k+l==0)continue;
							tl-=(n-m);
							tk+=(n-m);
							if(tl<0||tk<0)continue;
							long long ks=finv[tk]*finv[tl]%mod;
							dp[i+1][j+m+n][m][n]=(dp[i+1][j+m+n][m][n]+dp[i][j][k][l]*ks)%mod;
						}
					}
				}
			}
		}
	}
	long long ret=0;
	for(int i=a;i<=a;i++){
		for(int j=0;j<=a;j++){
			for(int k=0;k<=a;k++){
				if(!dp[i][a][j][k])continue;

				// printf("%d %d %d: %lld\n",i,j,k,dp[i][a][j][k]);
				long long ks=finv[j]*finv[k]%mod;
				ret=(ret+dp[i][a][j][k]*ks)%mod;
			}
		}
	}

	ret=ret*fact[a]%mod;
	printf("%lld\n",ret);
}