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# (Mono-mcs 6.8.0.105) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
class Program
{
	static void Main(string[] args)
	{
		string[]str = Console.ReadLine().Split();
		int N = int.Parse(str[0]);
		int M = int.Parse(str[1]);
		int K = int.Parse(str[2]);
		long[]D = new long[K+2];
		D[0] = M;
		long mod = 998244353;
		long[]DA = new long[200010];
		long[]DB = new long[200010];
		DA[0] = 1;
		DB[0] = 1;
		for(var i=1;i<200010;i++){
			DA[i] = DA[i-1]*i%mod;
			DB[i] = DB[i-1]*Mpow(i,mod-2,mod)%mod;
		}
		long ans = 0;
		for(var i=0;i<=K;i++){
          	//Console.Error.WriteLine(Mpow(M-1,N-1-i,mod)+" "+DA[N]+" "+DB[N-i]+" "+DB[i]);
			ans += M * Mpow(M-1,N-1-i,mod) % mod * DA[N-1] % mod * DB[N-1-i] % mod * DB[i] % mod;
			ans %= mod;
		}
		Console.WriteLine(ans%mod);
	}
	
	public static long Mpow(long x,long n,long m)
	{
		long ans = 1;
		while(n > 0){
			if(n % 2 == 1){
				ans = ans * x % m;
				n -= 1;
			}
			x = x * x % m;
			n /= 2;
		}
		return ans;
	}
}