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
/*
读题不规范，爆零两行泪。
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
const int N=300,M=300;
int n,m,mod;
int comb[N+1][N+1];
int dp[N+1][M+2];
int Sum[N+1][M+2];
int main(){
	cin>>n>>m>>mod;
	comb[0][0]=1;
	for(int i=1;i<=n;i++)for(int j=0;j<=i;j++)comb[i][j]=((j?comb[i-1][j-1]:0)+(j<i?comb[i-1][j]:0))%mod;
	for(int i=1;i<=m+1;i++)Sum[0][i]=(Sum[0][i-1]+(dp[0][i]=1))%mod;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			for(int k=1;k<=i;k++)(dp[i][j]+=1ll*dp[i-k][j]*comb[i-1][k-1]%mod*(Sum[k-1][m+1]-Sum[k-1][j])%mod)%=mod;
			Sum[i][j]=(Sum[i][j-1]+dp[i][j])%mod;
//			printf("dp[%d][%d]=%d\n",i,j,dp[i][j]);
		}
		Sum[i][m+1]=Sum[i][m];
	}
	cout<<(dp[n][1]+mod)%mod;
	return 0;
}