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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

// practice with kaiboy
import java.io.*;
import java.util.*;

public class Main extends PrintWriter {
	Main() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		Main o = new Main(); o.main(); o.flush();
	}

	static final int MD = 998244353;
	long power(long a, int k) {
		long p = 1;
		while (k > 0) {
			if ((k & 1) == 1)
				p = p * a % MD;
			a = a * a % MD;
			k >>= 1;
		}
		return p;
	}
	int[] ff, gg;
	void init(int n) {
		ff = new int[n];
		gg = new int[n];
		for (int f = 1, i = 0; i < n; i++) {
			gg[i] = (int) power(ff[i] = f, MD - 2);
			f = (int) ((long) f * (i + 1) % MD);
		}
	}
	long ch(int n, int k) {
		return (long) ff[n] * gg[k] % MD * gg[n - k] % MD;
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		init(n);
		long ans = 0;
		for (int h = 0; h <= k; h++)
			ans = (ans + ch(n - 1, h) * m % MD * power(m - 1, n - 1 - h)) % MD;
		println(ans);
	}
}