Find the number of the possible tuples of sequences (A_0,A_1,...,A_N) that satisfy all of the following conditions, modulo M:

* For every i (0\leq i\leq N), A_i is a sequence of length i consisting of integers between 1 and K (inclusive);
* For every i (1\leq i\leq N), A_{i-1} is a subsequence of A_i, that is, there exists 1\leq x_i\leq i such that the removal of the x_i-th element of A_i would result in a sequence equal to A_{i-1};
* For every i (1\leq i\leq N), A_i is lexicographically larger than A_{i-1}.

Constraints

* 1 \leq N,K \leq 300
* 2 \leq M \leq 10^9
* N, K and M are integers.

Input

Input is given from Standard Input in the following format:


N K M


Output

Print the number of the possible tuples of sequences (A_0,A_1,...,A_N), modulo M.

Examples

Input

2 2 100


Output

5


Input

4 3 999999999


Output

358


Input

150 150 998244353


Output

186248260
#include <bits/stdc++.h>
#define RI register int
typedef long long LL;

#define FILEIO(name) freopen(name".in", "r", stdin), freopen(name".out", "w", stdout);

using namespace std;

int f[305][305][305];

int main() {

#ifdef LOCAL
  FILEIO("a");
#endif

  int n, k, mod; cin >> n >> k >> mod;
  f[0][1][1] = 1;
  for (RI i = 0; i <= n; ++i)
    for (RI j = 1; j <= k; ++j)
      for (RI l = n + 1; l; --l) {
        if (l == 1) (f[i][j + 1][i + 1] += f[i][j][l]) %= mod;
        else (f[i][j][l - 1] += f[i][j][l]) %= mod;
        (f[i + 1][j][l] += 1ll * f[i][j][l] * l % mod) %= mod;
      }
  cout << f[n][k][1] << endl;

  cerr << (double)(clock()) / CLOCKS_PER_SEC * 1000 << "ms" << endl;

  return 0;
}

// created by Daniel yuan