Problem: Ember and Storm play a game. First, Ember picks a labelled tree T of n vertices, such that the degree of every vertex is at most d. Then, Storm picks two distinct vertices u and v in this tree and writes down the labels of the vertices in the path from u to v in a sequence a1, a2... ak. Finally, Ember picks any index i (1 ≤ i < k) in the array. Now he performs one of the following two operations exactly once:

  * flip the subrange [i + 1, k] and add ai to it. After this, the sequence becomes a1, ... ai, ak + ai, ak - 1 + ai, ... ai + 1 + ai
  * negate the subrange [i + 1, k] and add ai to it. i.e., the array becomes a1, ... ai, - ai + 1 + ai, - ai + 2 + ai, ... - ak + ai



Ember wins if the array is monotonically increasing or decreasing after this. Otherwise Storm wins.

The game can be described by the tuple (T, u, v, i, op) where op is «flip» or «negate» depending on the action Ember chose in the last turn. Find the number of tuples that can occur if Ember and Storm play optimally. When they play optimally, if there are multiple moves by which they are guaranteed to win, then they may play any of the winning moves. Otherwise, if someone loses no matter what they play, then they may play any of the possible moves.

Report the answer modulo m.

Input

The input consists of a single line containing three integers n, d and m (2 ≤ n ≤ 200, 1 ≤ d < n, 1 ≤ m ≤ 2·109).

Output

Print a single number — the number of possible tuples if Ember and Storm play as described, modulo m.

Examples

Input

2 1 1000000007


Output

4


Input

3 1 250


Output

0


Input

3 2 100


Output

36

Note

In the first sample case, there is only one possible tree. There are two possible paths, 1 to 2 and 2 to 1. For both paths, i can only be 1, and op can take both possibilities. Therefore, the answer is 4.

In the second sample, there are no possible trees.

In the third sample, there are three possible trees. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0},  {0, -1},
                        {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <typename t, typename t1>
t min1(t a, t1 b) {
  return a < b ? a : b;
}
template <typename t, typename... arg>
t min1(t a, arg... arr) {
  return min1(a, min1(arr...));
}
template <typename t, typename t1>
t max1(t a, t1 b) {
  return a > b ? a : b;
}
template <typename t, typename... arg>
t max1(t a, arg... arr) {
  return max1(a, max1(arr...));
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / max(abs1(a), abs1(b)) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 0) return 0;
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitnum(long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitmaxl(unsigned int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
inline int bitmaxl(long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
  if (mod == -1) mod = 1000000007;
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <class t>
void output1(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) cerr << arr[i] << ' ';
  cerr << endl;
}
template <class t>
void output2(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) output1(arr[i]);
}
const int maxn = 210;
long long mono[maxn][maxn], sum_mono[maxn][maxn], multi_mono[maxn][maxn];
long long c[maxn][maxn];
int n, d, mod;
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d%d%d", &n, &d, &mod);
  for (int i = 0; i < maxn; i++) {
    c[i][0] = 1;
    for (int j = 1; j < maxn; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      c[i][j] %= mod;
    }
  }
  mono[1][0] = 1;
  for (int i = 0; i <= d; i++) sum_mono[1][i] = 1;
  multi_mono[0][0] = 1;
  multi_mono[1][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= i - 1; j++) {
      for (int rd = 0; rd <= d - 1; rd++) {
        mono[i][rd + 1] +=
            mono[i - j][rd] * c[i - 2][j - 1] % mod * sum_mono[j][d - 1];
        mono[i][rd + 1] %= mod;
      }
      for (int rd = 1; rd <= d - 1; rd++) {
        multi_mono[i][rd + 1] += multi_mono[j][rd] * multi_mono[i - j][1] %
                                 mod * c[i - 1][i - j - 1];
        multi_mono[i][rd + 1] %= mod;
      }
    }
    sum_mono[i][0] = mono[i][0];
    for (int rd = 1; rd <= d; rd++) {
      sum_mono[i][rd] = mono[i][rd] + sum_mono[i][rd - 1];
      sum_mono[i][rd] %= mod;
    }
    multi_mono[i][1] = sum_mono[i][d - 1];
  }
  long long ans = 0;
  for (int i = 0; i < n; i++)
    for (int rd_small = 0; rd_small <= d; rd_small++) {
      ans += multi_mono[i][rd_small] * multi_mono[n - i - 1][0];
      ans %= mod;
      for (int rd_big = 2; rd_big + rd_small <= d; rd_big++) {
        ans += multi_mono[i][rd_small] * multi_mono[n - i - 1][rd_big];
        ans %= mod;
      }
    }
  cout << ans * 2 * (n - 1) % mod * n % mod << endl;
  return 0;
}