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 static System.Math;
using static System.Console;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace AtCoder
{
	class Program
	{
		static long mod = 998244353;
		static long[] fac, finv, inv;

		static void Main(string[] args)
		{
			var firstLine = ReadLine().Split(' ').Select(value => long.Parse(value)).ToArray();
			var n = firstLine[0];
			var m = firstLine[1];
			var k = firstLine[2];

			CombinationInitialize(n);

			long count = 0;
			for(long i = n - 1 - k; i <= n - 1; i++)
			{
				count += ModPow(m - 1, i, mod) * Combination(n - 1, n - 1 - i);
				count %= mod;
			}

			WriteLine(count * m % mod);
		}

		private static long ModPow(long a, long n, long mod)
		{
			long answer = 1;

			while(n > 0)
			{
				if((n & 1) == 1)
				{
					answer = answer * a % mod;
				}
				a = a * a % mod;
				n >>= 1;
			}

			return answer;
		}

		private static long Combination(long n, long k)
		{
			if(n < k)
			{
				return 0;
			}
			if(n < 0 || k < 0)
			{
				return 0;
			}

			return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
		}

		private static void CombinationInitialize(long n)
		{
			fac = new long[n + 1];
			finv = new long[n + 1];
			inv = new long[n + 1];

			fac[0] = fac[1] = 1;
			finv[0] = finv[1] = 1;
			inv[1] = 1;

			for(long i = 2; i <= n; i++)
			{
				fac[i] = fac[i - 1] * i % mod;
				inv[i] = mod - inv[mod % i] * (mod / i) % mod;
				finv[i] = finv[i - 1] * inv[i] % mod;
			}
		}
	}
}