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<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

template<unsigned MOD> struct ModInt {
    static const unsigned static_MOD = MOD;
    unsigned x;
    void undef() { x = (unsigned)-1; }
    bool isnan() const { return x == (unsigned)-1; }
    inline int geti() const { return (int)x; }
    ModInt() { x = 0; }
    ModInt(const ModInt &y) { x = y.x; }
    ModInt(int y) { if (y<0 || (int)MOD<=y) y %= (int)MOD; if (y<0) y += MOD; x=y; }
    ModInt(unsigned y) { if (MOD<=y) x = y % MOD; else x = y; }
    ModInt(long long y) { if (y<0 || MOD<=y) y %= MOD; if (y<0) y += MOD; x=y; }
    ModInt(unsigned long long y) { if (MOD<=y) x = y % MOD; else x = y; }
    ModInt &operator+=(const ModInt y) { if ((x += y.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(const ModInt y) { if ((x -= y.x) & (1u<<31)) x += MOD; return *this; }
    ModInt &operator*=(const ModInt y) { x = (unsigned long long)x * y.x % MOD; return *this; }
    ModInt &operator/=(const ModInt y) { x = (unsigned long long)x * y.inv().x % MOD; return *this; }
    ModInt operator-() const { return (x ? MOD-x: 0); }

    ModInt inv() const {
	unsigned a = MOD, b = x; int u = 0, v = 1;
	while (b) {
	    int t = a / b;
	    a -= t * b; swap(a, b);
	    u -= t * v; swap(u, v);
	}
	if (u < 0) u += MOD;
	return ModInt(u);
    }
    ModInt pow(long long y) const {
	ModInt b = *this, r = 1;
	if (y < 0) { b = b.inv(); y = -y; }
	for (; y; y>>=1) {
	    if (y&1) r *= b;
	    b *= b;
	}
	return r;
    }
    friend ModInt operator+(ModInt x, const ModInt y) { return x += y; }
    friend ModInt operator-(ModInt x, const ModInt y) { return x -= y; }
    friend ModInt operator*(ModInt x, const ModInt y) { return x *= y; }
    friend ModInt operator/(ModInt x, const ModInt y) { return x *= y.inv(); }
    friend bool operator<(const ModInt x, const ModInt y) { return x.x < y.x; }
    friend bool operator==(const ModInt x, const ModInt y) { return x.x == y.x; }
    friend bool operator!=(const ModInt x, const ModInt y) { return x.x != y.x; }
};

const LL MOD = 1000000007;
typedef ModInt<MOD> Mint;
const int MAX = 1000000;
Mint inv[MAX+1], fact[MAX+1], fact_inv[MAX+1];

void init() {
    inv[1] = 1;
    for (int i=2; i<=MAX; i++) inv[i] = inv[MOD%i] * (MOD-MOD/i);

    fact[0] = fact_inv[0] = 1;
    for (int i=1; i<=MAX; i++) {
	fact[i] = fact[i-1] * i;
	fact_inv[i] = fact_inv[i-1] * inv[i];
    }
}

Mint nCk(int n, int k) {
    return fact[n] * fact_inv[k] * fact_inv[n-k];
}

int N;
Mint dp[51][101];

void MAIN() {
    init();

    scanf("%d", &N);

    dp[1][50] = N; // 1;
    dp[1][49] = N; // -1;
    Mint ans = 0;

    for (int t=1; t<=51; t++) {
	ans += dp[N][50];
	for (int i=N; i>=0; i--) {
	    for (int j=-50; j<=50; j++) if (dp[i][j+50].x) {
		REP (a, N+1) REP (b, N+1) if ((a|b) && i+a+b <= N) {
		    int m = a+b;
		    if (-m <= j && j <= m && (j+m)%2 == 0) {
			dp[i+m][(j+a-b)/2+50] += dp[i][j+50] * nCk(N-i, a) * nCk(N-i-a, b);
		    }
		}
		dp[i][j+50] = 0;
	    }
	}
    }

    printf("%d\n", ans.geti());
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}

