// http://noi.ac/problem/37

#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

typedef unsigned long long ull;
const int MAX_N = 5001;

ull mod[MAX_N];  // n! % p的结果
// 如果m > n，就是m的n次方模p的结果
// a的b次方模p
// 假设b = 11，即1011，那么pow(a, b) = pow(a, 8) * pow(a, 2) * pow(a, 1)
ull fast_pow_mod(ull a, ull b, ull p) {
    ull res = 1;
    ull cur = a % p;
    while (b > 0) {
        if (b & 1) {
            res = res * cur % p;
        }
        cur = cur * cur % p;
        b >>= 1;
    }
    return res;
}

int simple(int n, int m, int p) {
    // 结尾最多有j个各不相同的涂色的方案数
    static ull cur_num[MAX_N];
    static ull prev_num[MAX_N];
    static ull suffix_sum[MAX_N];

    ull res = 0;
    for (int i = 1; i <= n; ++i) {
        int max_j = min(m - 1, i);
        for (int j = 1; j <= max_j; ++j) {
            if (i == 1) {
                cur_num[j] = m;
            }
            else {
                cur_num[j] = prev_num[j-1] * (m - j + 1) % p;
                cur_num[j] = (cur_num[j] + suffix_sum[j]) % p;
            }

            if (i == n) {
                res = (res + cur_num[j]) % p;
            }
        }

        if (i < n) {
            for (int j = 1; j <= max_j; ++j) {
                prev_num[j] = cur_num[j];
            }
            for (int j = max_j; j >= 1; --j) {
                suffix_sum[j] = (j == max_j) ? cur_num[j] : ((suffix_sum[j+1] + cur_num[j]) % p);
            }
        }
    }
    return res;
}

int main() {
    int n, m, p;

    while (cin >>n >>m >>p) {
        if (m == 1) {
            cout <<1 <<endl;
        }
        else if (m > n) {
            cout <<fast_pow_mod(m, n, p) <<endl;
        }
        else {
            cout <<(simple(n, m, p) % p) <<endl;
        }
    }

    return 0;
}