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;
const int N = 311;
#define fr(i,a,b) for(int i = a;i <= b; ++i)
#define nfr(i,a,b) for(int i = a;i >= b; --i)

int n,m,p;
int f[N][N];
int sum[N];
int c[N][N];

inline void add(int &x,int y){
   x += y; if(x >= p)x -= p;
}

int main(){
   scanf("%d%d%d",&n,&m,&p);
   memset(f,0,sizeof f);
   
   fr(i,0,n)c[i][0] = 1;
   fr(i,1,n){
      fr(j,1,n){
         c[i][j] = (c[i-1][j]+c[i-1][j-1])%p;
      }
   }
   
   fr(i,0,m)f[i][1] = 1;
   fr(i,2,n+1){
      memset(sum,0,sizeof sum);
      nfr(j,m-1,0){
         fr(k,1,i-1)add(sum[k],f[j+1][k]);
         fr(k,1,i-1){
            add(f[j][i],1LL*f[j][i-k]*sum[k]%p*c[i-2][k-1]%p);
         }
      }
   }
   printf("%d\n",f[0][n+1]);
   return 0;
}
