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 = 3E2;
int n, k, p;
int dp[N + 2], c[N + 2][N + 2], sum[N + 2];
void inc(int &x, int y) {(x += y) >= p ? x -= p : 0;}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> k >> p;
    for (int i = 0; i <= N + 1; ++i) {
        c[i][0] = 1;
        for (int j = 1; j <= i; ++j)
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % p;
    }
    for (int r = k; r >= 0; --r) {
        memset(dp, 0, sizeof(dp));
        dp[1] = 1;
        for (int i = 1; i <= n; ++i)
            for (int s = 1; i + s <= n + 1; ++s)
                inc(dp[i + s], 1LL * dp[i] * sum[s] % p * c[i + s - 2][s - 1] % p);
        for (int s = 1; s <= n + 1; ++s)
            inc(sum[s], dp[s]);
    }
    cout << dp[n + 1] << endl;
    return 0;
}