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 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main{
	static Scanner sc = new Scanner(System.in);
	static long mod = 998244353;
	public static void main(String[] args) {
		int n = sc.nextInt(), m = sc.nextInt(), k = sc.nextInt();
		initComb();
		long ans = 0;
		for(int i=0;i<=k;i++) {
			ans += comb(n-1, i) * pow(m-1, n-i-1) % mod;
			ans %= mod;
		}
		ans = ans * m % mod;//先頭の塗り方
		System.out.println(ans);
	}
	static long fact[] = new long[200200];
	static long revFact[] = new long[200200];
	
	static void initComb() {
		fact[0] = 1;
		for(int i=1;i<200200;i++) {
			fact[i] = fact[i-1] * i % mod;
		}
		revFact[200199] = div(1, fact[200199]); 
		for(int i=200199;i>=1;i--) {
			revFact[i-1] = revFact[i] * i % mod;
		}
	}
	
	static long div(long x, long y) {
		return x * pow(y, mod - 2) % mod;
	}
	
	static long pow(long a, long x) {
		if(x == 0)return 1;
		if(x % 2 == 1) return a * pow(a, x-1) % mod;
		long root = pow(a, x / 2);
		return root * root % mod;
	}
	
	static long comb(int x, int y) {
		if( x < y )return 0;
		long up = fact[x];
		long down = fact[x - y] * fact[y] % mod;
		return div(up, down);
	}
}