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;

typedef long long ll;
const int maxn = 55, mod = 1e9 + 7;

inline int gi()
{
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	int sum = 0;
	while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

int add(int x, int y) {return x + y < mod ? x + y : x + y - mod;}

int n, f[maxn][maxn], C[maxn][maxn];

int main()
{
	n = gi();
	f[1][0] = f[1][1] = 1;
	for (int i = 0; i <= n; ++i) C[i][0] = 1;
	for (int i = 1; i <= n; ++i)
		for (int j = C[i][0] = 1; j <= i; ++j) C[i][j] = add(C[i - 1][j - 1], C[i - 1][j]);

	for (int i = 1; i < n; ++i)
		for (int j = 0; j <= i; ++j)
			if (f[i][j])
				for (int x = 0; x + i <= n; ++x)
					for (int y = 0; x + y + i <= n; ++y)
						if (x + y && x + y >= j && ((x + y - j) & 1) == 0)
							f[i + x + y][abs((x + y - j) / 2 + j - x)] = (f[i + x + y][abs((x + y - j) / 2 + j - x)] + (ll)f[i][j] * C[i + x + y][x + y] % mod * C[x + y][x]) % mod;

	printf("%d\n", f[n][0]);
	
	return 0;
}
