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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define int long long
#define MAXN 310
int n,k,MOD,f[MAXN][MAXN][MAXN];
inline void add(int&a,int b){a=(a+b)%MOD;}
signed main(){
	ios_base::sync_with_stdio(false);
	cin>>n>>k>>MOD;
	f[0][1][0]=1;
	for(int i=0;i<=n;++i)
		for(int v=1;v<=k;++v)
			for(int p=i;~p;--p){
				if(p)add(f[i][v][p-1],f[i][v][p]);
				else add(f[i][v+1][i],f[i][v][p]);
				add(f[i+1][v][p],f[i][v][p]*(p+1)%MOD);
			}
	cout<<f[n][k+1][n]<<endl;
	return 0;
}