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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
const int maxn=300+7;
int n,K,MOD;
int now[maxn];
int C[maxn][maxn];
int f[maxn][maxn];
int g[maxn][maxn];
int main(){
	scanf("%d%d%d",&n,&K,&MOD);
	for(int i=0;i<=K;++i)
		for(int j=C[i][0]=1;j<=i;++j)
			C[i][j]=(C[i-1][j-1]+C[i-1][j])%MOD;
	for(int i=0;i<n;++i){
		for(int k=1;k<=i+1;++k)now[k]=k;
		for(int j=1,sum=0;j<=n-i;++j,sum=0){
			for(int k=1;k<=i+1;++k)
				sum=(sum+now[k])%MOD,now[k]=sum*1ll*k%MOD;
			g[i][j]=sum;
		}
	}
	f[0][0]=1;
	for(int i=0;i<n;++i)
		for(int j=0;j<K;++j){
			if(!f[i][j])continue;
			const int F=f[i][j];
			for(int k=1;k<=n-i;++k)
				f[i+k][j+1]=(f[i+k][j+1]+F*1ll*g[i][k])%MOD;
		}
	int ans=0;
	for(int i=1;i<=K;++i)
		ans=(ans+f[n][i]*1ll*C[K][i])%MOD;
	printf("%d\n",ans);
	return 0;
}