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)

#include <bits/stdc++.h>

using namespace std;

const int P = 998244353;

long long power(long long x, long long t) {
	long long ret = 1;
	for(; t; t >>= 1, x = x * x % P) {
		if(t & 1) {
			ret = ret * x % P;
		}
	}
	return ret;
}

int main() {
	int n, m, k;
	cin >> n >> m >> k;

	vector<long long> fac(n + 1);
	fac[0] = 1;
	for(int i = 1; i <= n; ++i) {
		fac[i] = fac[i - 1] * i % P;
	}

	vector<long long> inv(n + 1);
	inv[n] = power(fac[n], P - 2);

	for(int i = n; i >= 1; --i) {
		inv[i - 1] = inv[i] * i % P;
	}

	auto C = [&] (int n, int m) {
		if(m > n) {
			return 0LL;
		}
		return fac[n] * inv[m] % P * inv[n - m] % P;
	};

	long long ans = 0;

	for(int i = n - k - 1; i < n; ++i) {
		ans = (ans + C(n - 1, i) * power(m - 1, i) % P * m % P) % P;
	}

	cout << ans << '\n';

	return 0;
}