Problem: A tree is a connected graph without cycles.

Two trees, consisting of n vertices each, are called isomorphic if there exists a permutation p: {1, ..., n} → {1, ..., n} such that the edge (u, v) is present in the first tree if and only if the edge (pu, pv) is present in the second tree.

Vertex of the tree is called internal if its degree is greater than or equal to two.

Count the number of different non-isomorphic trees, consisting of n vertices, such that the degree of each internal vertex is exactly d. Print the answer over the given prime modulo mod.

Input

The single line of the input contains three integers n, d and mod (1 ≤ n ≤ 1000, 2 ≤ d ≤ 10, 108 ≤ mod ≤ 109) — the number of vertices in the tree, the degree of internal vertices and the prime modulo.

Output

Print the number of trees over the modulo mod.

Examples

Input

5 2 433416647


Output

1


Input

10 3 409693891


Output

2


Input

65 4 177545087


Output

910726

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
static const long long MAXN = 1005;
static const long long MAXD = 11;
static const long long MAX_MODULUS = 1e9 + 7;
template <typename T>
inline T min(const T a, const T b) {
  return a < b ? a : b;
}
long long n, d, mod;
inline long long fpow(long long base, long long exp) {
  long long ans = 1;
  for (; exp; exp >>= 1) {
    if (exp & 1) (ans *= base) %= mod;
    base = (long long)base * base % mod;
  }
  return ans;
}
inline long long _inv(const long long x) { return fpow(x, mod - 2); }
long long inv[MAXD];
inline void calc_inversions() {
  for (long long i = 1; i < MAXD; ++i) inv[i] = _inv(i);
}
inline long long binom_2(const long long x) {
  return (long long)x * (x - 1) % mod * inv[2] % mod;
}
long long f[MAXN][MAXD][MAXN] = {{{0}}};
signed main() {
  scanf("%I64d%I64d%I64d", &n, &d, &mod);
  if (n <= 2) {
    puts("1");
    return 0;
  }
  calc_inversions();
  for (long long i = 0; i <= n; ++i) f[0][0][i] = 1;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= d; ++j)
      for (long long k = 1; k <= min(i, (n + 1) / 2); ++k) {
        long long binom_begin = (k == 1 ? 1 : f[k - 1][d - 1][k - 1]);
        long long binom_val = 1;
        for (long long t = 0; t <= i / k && t <= j; ++t) {
          f[i][j][k] =
              ((long long)f[i][j][k] +
               binom_val * f[i - k * t][j - t][min(i - k * t, k - 1)]) %
              mod;
          binom_val =
              (long long)binom_val * (binom_begin + t) % mod * inv[t + 1] % mod;
        }
      }
  long long ans =
      (f[n - 1][d][(n - 1) / 2] +
       ((n & 1) ? 0 : binom_2(f[n / 2 - 1][d - 1][(n - 1) / 2] + 1))) %
      mod;
  printf("%I64d\n", ans);
  return 0;
}