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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int Mod;
int dp[310][310];
int C[310][310];
int Dp[310][310];
int f(int,int);
int F(int n,int k)
{
	if(k<0)return 0;
	if(Dp[n][k]!=-1)return Dp[n][k];
	return Dp[n][k]=(F(n,k-1)+f(n,k))%Mod;
}
int f(int n,int k)
{
	if(k<0)return 0;
	if(n==0)return 1;
	if(dp[n][k]!=-1)return dp[n][k];
	dp[n][k]=0;
	for(int i=1;i<=n;i++)
		dp[n][k]=(dp[n][k]+(long long)F(i-1,k-1)*f(n-i,k)%Mod*C[n-1][i-1])%Mod;
	return dp[n][k];
}
int main()
{
	int n,k;
	scanf("%d %d %d",&n,&k,&Mod);
	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;
	}
	memset(dp,-1,sizeof(dp));
	memset(Dp,-1,sizeof(Dp));
	printf("%d\n",f(n,k));
	return 0;
}