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 <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N;
int dp[55][105],inv[55],invfac[55],fac[55];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    read(N);
    inv[1] = 1;
    for(int i = 2 ; i <= N ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    fac[0] = invfac[0] = 1;
    for(int i = 1 ; i <= N ; ++i) {
	fac[i] = mul(fac[i - 1],i);
	invfac[i] = mul(invfac[i - 1],inv[i]);
    }
    dp[1][N] = 1;
    dp[1][N - 1] = 1;
    for(int i = 1 ; i <= N ; ++i) {
	for(int j = -i ; j <= i ; ++j) {
	    for(int k = max(1,abs(j)) ; k <= N ; ++k) {
		if(i + k > N) break;
		for(int h = 0 ; h <= k ; ++h) {
		    if((j + 2 * h - k) % 2) continue;
		    int val = (j + 2 * h - k) / 2;
		    if(val < -N || val > N) continue;
		    update(dp[i + k][val + N],mul(dp[i][j + N],mul(C(i + k,k),C(k,h))));
		}
	    }
	}
    }
    out(dp[N][N]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}
