Problem:
Score : 500 points

Problem Statement

There are N blocks arranged in a row. Let us paint these blocks.

We will consider two ways to paint the blocks different if and only if there is a block painted in different colors in those two ways.

Find the number of ways to paint the blocks under the following conditions:

For each block, use one of the M colors, Color 1 through Color M, to paint it. It is not mandatory to use all the colors.

There may be at most K pairs of adjacent blocks that are painted in the same color.

Since the count may be enormous, print it modulo 998244353.

Constraints

All values in input are integers.

1 \leq N, M \leq 2 \times 10^5

0 \leq K \leq N - 1

Input

Input is given from Standard Input in the following format:

N M K

Output

Print the answer.

Sample Input 1

3 2 1

Sample Output 1

6

The following ways to paint the blocks satisfy the conditions: 112, 121, 122, 211, 212, and 221. Here, digits represent the colors of the blocks.

Sample Input 2

100 100 0

Sample Output 2

73074801

Sample Input 3

60522 114575 7559

Sample Output 3

479519525

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#ifdef MY_LOCAL
#define MY_NAMESPACE(ns) namespace ns {
#define MY_NAMESPACE_    }
#define MY_DEBUG(s)      s
#else
#define MY_NAMESPACE(ns) 
#define MY_NAMESPACE_    
#define MY_DEBUG(s)      
#endif

#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <string>
#include <tuple>
#include <unordered_map>
#include <utility>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define EPS (1e-7)
#define IINF ((int)1e9)


MY_NAMESPACE(testbed)
ll modpow(ll base, ll exponent, ll mod)
{
	ll retval = 1;
	while (exponent > 0)
	{
		if (exponent % 2)
			retval = (retval * base) % mod;
		exponent /= 2;
		base = (base * base) % mod;
	}
	return retval;
}

class ModComb
{
public:
	ModComb(int max_n, ll mod) : m_mod(mod)  // mod and any i (1<i<=max_n) must be mutually prime. in reality mod must be a prime number and mod>max_n
	{
		m_fac.resize(max_n + 1);
		m_inv.resize(max_n + 1);
		m_fin.resize(max_n + 1);
		m_fac[0] = 1;
		m_fac[1] = 1;
		m_inv[0] = 0;  // dummy
		m_inv[1] = 1;
		m_fin[0] = 1;
		m_fin[1] = 1;
		for (int i = 2; i <= max_n; i++)
		{
			m_fac[i] = m_fac[i - 1] * i % mod;
			m_inv[i] = mod - m_inv[mod % i] * (mod / i) % mod;
			m_fin[i] = m_fin[i - 1] * m_inv[i] % mod;
		}
	}

	ll v(int n, int k)
	{
		if (n < 0 || k < 0)
			return 0;
		if (n < k)
			return 0;
		return m_fac[n] * (m_fin[k] * m_fin[n - k] % m_mod) % m_mod;
	}

private:
	ll m_mod;
	vector<ll> m_fac;  // factorial
	vector<ll> m_inv;  // inverse element
	vector<ll> m_fin;  // factorial of inverse
};

int main()
{
	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	ll const modval = 998244353;

	int n, m, k;
	cin >> n >> m >> k;

	ModComb modcomb(n, modval);

	ll res = 0;
	for (int this_k = 0; this_k <= k; this_k++)
	{
		ll this_res = modcomb.v(n - 1, this_k);
		this_res = this_res * m % modval;
		this_res = this_res * modpow((ll)m - 1, (ll)n - 1 - (ll)this_k, modval) % modval;
		res = (res + this_res) % modval;
	}
	cout << res << "\n";
	return 0;
}
MY_NAMESPACE_