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 N = 51;
const int mod = 1e9 + 7;

typedef long long LL;

inline int Pow(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;
}

int frac[N], inv[N];

inline int C(int n, int r) {
    return (LL)frac[n] * inv[r] % mod * inv[n - r] % mod;
}

int dp[N][N]; // 已经挂了i个点，最后一层有j个点有奇数个儿子，还未考虑儿子的挂法

int main() {
    int n; scanf("%d", &n), frac[0] = inv[0] = 1;
    for (int i = 1; i <= n; i++) frac[i] = (LL)frac[i - 1] * i % mod, inv[i] = Pow(frac[i], mod - 2);
    dp[1][1] = dp[1][0] = n;
    for (int i = 1; i < n; i++)
        for (int j = 1; j <= n - i; j++) // 当前这一层的点数
            for (int k = 0; k <= j; k++) { // 上一层奇数个儿子的点的数量
                if ((j - k) & 1) continue;
                int delta = (j - k) / 2; // 未考虑下一层时，这一层有多少个点与父亲符号不同
                for (int t = 0; t <= j; t++) { // 实际有多少个符号不同
                    int tmp = abs(t - delta); // 这一层有奇数个儿子的点的数量
                    dp[i + j][tmp] = (dp[i + j][tmp] + (LL)dp[i][k] * C(n - i, j) % mod * C(j, t)) % mod;
                }
            }
    printf("%d\n", dp[n][0]);
    return 0;
}