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>
using namespace std;
const int MAXN = 105;
const int P = 1e9 + 7;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
int dp[MAXN][MAXN];
int fac[MAXN], inv[MAXN];
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int power(int x, int y) {
	if (y == 0) return 1;
	int tmp = power(x, y / 2);
	if (y % 2 == 0) return 1ll * tmp * tmp % P;
	else return 1ll * tmp * tmp % P * x % P;
}
int main() {
	int n; read(n);
	fac[0] = inv[0] = 1;
	for (int i = 1; i <= n; i++) {
		fac[i] = 1ll * fac[i - 1] * i % P;
		inv[i] = power(fac[i], P - 2);
	}
	dp[n - 1][n] = dp[n - 1][n + 1] = 1;
	for (int i = n - 1; i >= 1; i--)
	for (int j = 0; j <= 2 * n; j++) {
		int r = j - n;
		for (int p = 0; p <= i; p++)
		for (int q = 0; p + q <= i; q++)
			if (p + q != 0 && (r + p - q) % 2 == 0 && r + p + q >= 0 && r - p - q <= 0) {
				update(dp[i - p - q][n + (r + p - q) / 2], 1ll * dp[i][j] * inv[p] % P * inv[q] % P);
			}
	}
	writeln(1ll * fac[n] * dp[0][n] % P);
	return 0;
}