Rikka with Nash Equilibrium

/****** 标程 *******/
const int N = 85;

int n, m, mod;
int dp[2][N][N];

void update(int &k1, LL k2) {
    k1 = (k1 + k2) % mod;
}

void Solve() {
    mem(dp, 0x00);
    dp[0][1][1] = n * m;

    int now = 0;
    for (int i = 2; i <= n * m; ++i) {
        int nxt = now ^ 1;
        mem(dp[nxt], 0x00);
        for (int j = 1; j <= n; ++j) {
            for (int k = 1; k <= m; ++k) if (dp[now][j][k]) {
                int k1 = dp[now][j][k];
                update(dp[nxt][j][k], 1ll * k1 * (j * k - i + 1));
                update(dp[nxt][j + 1][k], 1ll * k1 * k * (n - j));
                update(dp[nxt][j][k + 1], 1ll * k1 * j * (m - k));
            }
        }
        now = nxt;
    }

    cout << dp[now][n][m] << endl;
}

int main()
{
    BEGIN() {
        cin >> n >> m >> mod;
        Solve();
    }
    return 0;
}