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
// I need some holiday
// But I don't wish that everyday is holiday
// Because I didn't bear the pain
// Because it wasn't a true freedom
//         ——『イヤホンと蝉時雨』·Orangestar

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef unsigned int UI;
typedef pair<int, int> pii;

#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define MP make_pair
#define fst first
#define snd second
#define SZ(u) ((int) (u).size())
#define ALL(u) (u).begin(), (u).end()

inline void proc_status()
{
	ifstream t("/proc/self/status");
	cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}

template<typename T> inline int chkmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }
template<typename T> inline int chkmax(T &a, const T &b) { return b > a ? a = b, 1 : 0; }
template<typename T> inline T sqr(const T &val) { return val * val; }

namespace fastIO
{
	const int MAX_BUFFER_SIZE = 1 << 16;

	char buffer[MAX_BUFFER_SIZE], *cur = buffer, *ed = buffer;

	inline char getc() { return cur == ed && (ed = (cur = buffer) + fread(buffer, 1, MAX_BUFFER_SIZE, stdin), cur == ed) ? EOF : *cur++; }
}
using fastIO::getc;

template<typename T> inline T read()
{
	register T sum(0), fg(1);
	register char ch(getc());
	for(; !isdigit(ch); ch = getc()) if(ch == '-') fg = -1;
	for(;  isdigit(ch); ch = getc()) sum = sum * 10 - '0' + ch;
	return sum * fg;
}

inline char *read_str(char *s)
{
	register char ch(getc());
	while(!isgraph(ch)) ch = getc();
	for(; isgraph(ch); ch = getc()) *s++ = ch;
	*s = '\0';
	return s;
}

// END tpl.

const int MOD = (int) 1e9 + 7;

inline int fpm(int x, int y)
{
	int res = 1;
	for(; y; y >>= 1, x = (LL) x * x % MOD) if(y & 1) res = (LL) res * x % MOD;
	return res;
}

inline int inv(int x) { assert(x); return fpm(x, MOD - 2); }

namespace MATH
{
	const int n = 50;

	int fac[n + 5], ifac[n + 5];

	inline void init()
	{
		fac[0] = 1;
		for(int i = 1; i <= n; ++i) fac[i] = (LL) fac[i - 1] * i % MOD;
		ifac[n] = inv(fac[n]);
		for(int i = n - 1; i >= 0; --i) ifac[i] = (LL) ifac[i + 1] * (i + 1) % MOD;
	}

	inline int C(int N, int M) { return N < 0 || M < 0 || N < M ? 0 : (LL) fac[N] * ifac[N - M] % MOD * ifac[M] % MOD; }
}
using MATH::C;

const int MAXN = 50;

int n;

inline void input()
{
	n = read<int>();
}

inline void solve()
{
	static int f[MAXN + 5][MAXN + 5];

	f[1][0] = f[1][1] = n;
	for(int i = 1; i < n; ++i)
		for(int j = 0; j <= n; ++j) if(f[i][j])
			for(int k = 1; i + k <= n; ++k) if(k >= j && (k - j) % 2 == 0)
			{
				int y = (k - j) / 2;
				for(int x = 0; x <= k; ++x) (f[i + k][abs(y - x)] += (LL) f[i][j] * C(n - i, k) % MOD * C(k, x) % MOD) %= MOD;
			}

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

int main()
{
#ifdef K_ON
	freopen("F.in", "r", stdin);
	freopen("F.out", "w", stdout);
#endif

	MATH::init();
	input();
	solve();

	return 0;
}