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;

#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define Debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef long double LD;
typedef unsigned int uint;
typedef pair <int, int> pii;
typedef unsigned long long uLL;

template <typename T> inline void Read(T &x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}

template <typename T> inline bool CheckMax(T &a, const T &b) {
  return a < b ? a = b, true : false;
}

template <typename T> inline bool CheckMin(T &a, const T &b) {
  return a > b ? a = b, true : false;
}

const int N = 105;
const int mod = 1e9 + 7;

int n, c[N][N], f[N][N];

int main() {
#ifdef wxh010910
  freopen("d.in", "r", stdin);
#endif
  Read(n);
  for (int i = 0; i <= n; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
    }
  }
  f[1][n] = f[1][n - 1] = n;
  for (int i = 1; i <= n; ++i) {
    for (int j = -n; j <= n; ++j) {
      if (f[i][j + n]) {
        for (int k = max(1, abs(j)); k <= n - i; ++k) {
          if (!(j + k & 1)) {
            for (int l = 0; l <= k; ++l) {
              int t = (j + k >> 1) - l;
              if (t >= -n && t <= n) {
                f[i + k][t + n] = (1LL * f[i][j + n] * c[n - i][k] % mod * c[k][l] + f[i + k][t + n]) % mod;
              }
            }
          }
        }
      }
    }
  }
  printf("%d\n", f[n][n]);
#ifdef wxh010910
  Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
  return 0;
}
