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>
#define rep(i,l,r) for (int i=l;i<=r;i++)
using namespace std;
typedef long long ll;
const int maxn=305;
ll f[maxn][maxn],g[maxn][maxn],c[maxn][maxn],fac[maxn],inv[maxn];
int n,k;
ll mm;
ll dfs(int i,int j);
ll getsum(int i,int j){
	if (~g[i][j]) return g[i][j];
	g[i][j]=0;
	rep(p,j,k) (g[i][j]+=dfs(i,p))%=mm;
	return g[i][j];
}
ll dfs(int i,int j){
	if (~f[i][j]) return f[i][j];
	if (i==1) return f[i][j]=1;
	f[i][j]=0;
	rep(len,1,i-1) {
		(f[i][j]+=getsum(len,j+1)*dfs(i-len,j)%mm*c[i-2][len-1]%mm)%=mm;
	}
	return f[i][j];
}
int main(){
	scanf("%d%d%lld",&n,&k,&mm);
	c[0][0]=1;
	rep(i,1,n) {
		c[i][0]=1; c[i][i]=1;
		rep(j,1,i-1) c[i][j]=(c[i-1][j]+c[i-1][j-1])%mm;
	}
	memset(f,-1,sizeof(f));
	memset(g,-1,sizeof(g));
	dfs(n+1,0);
	printf("%lld\n",f[n+1][0]);
	return 0;
}