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>
using namespace std;
#define ll long long 
int n,k;
ll mod;
ll f[330][330],s[330][330];
ll C[330][330];
void sol()
{
	for (int x=k;x>=0;--x) for (int size=1;size<=n+1;++size)
	{
		ll sss=0;
		for (int ch=1;ch<size;++ch)
		{
			ll doe=C[size-2][ch-1]*f[x][size-ch]%mod;
			ll now=(doe*s[x+1][ch]%mod+mod)%mod;
			sss=(sss+now)%mod;
		}
		if (size==1) sss=1;
		f[x][size]=sss;
		s[x][size]=(s[x+1][size]+f[x][size])%mod;

	}
}
int main()
{
//	freopen("E.in","r",stdin);
//	freopen("E.out","w",stdout);

	memset(f,-1,sizeof(f));
	scanf("%d%d%lld",&n,&k,&mod);
	C[0][0]=1;
	for (int i=1;i<=310;++i)
	{
		C[i][0]=C[i][i]=1;
		for (int j=1;j<i;++j)
			C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
	}
	
//	cerr<<"end_init"<<endl;
	sol();
	printf("%lld\n",f[0][n+1]);

	return 0;
}