Problem:
Score : 1200 points

Problem Statement

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.

Sample Input 1

2 2 100

Sample Output 1

5

Five tuples below satisfy the conditions:

(),(1),(1,1)

(),(1),(1,2)

(),(1),(2,1)

(),(2),(2,1)

(),(2),(2,2)

Sample Input 2

4 3 999999999

Sample Output 2

358

Sample Input 3

150 150 998244353

Sample Output 3

186248260

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 305;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
int n, m, P;
int dp[MAXN][MAXN], coef[MAXN][MAXN];
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int main() {
	read(n), read(m), read(P);
	dp[0][0] = coef[0][0] = 1;
	for (int i = 0; i <= n; i++)
	for (int j = 0; j <= n; j++) {
		if (i != 0) update(coef[i][j], coef[i - 1][j]);
		if (j != 0) update(coef[i][j], 1ll * coef[i][j - 1] * (i + 1) % P);
	}
	for (int i = 1; i <= m; i++)
	for (int j = 0; j <= n; j++) {
		for (int k = 0; k <= j; k++)
			update(dp[i][j], 1ll * coef[j - k][k] * dp[i - 1][j - k] % P);
	}
	writeln(dp[m][n]);
	return 0;
}