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 Ruby (2.7.1) (judge verdict: Accepted)

class BinomialCoefficients
  attr_reader :mod
  attr_accessor :fac, :finv, :inv

  def initialize(max, mod)
    @mod  = mod
    @fac  = Array.new(max)
    @finv = Array.new(max)
    @inv  = Array.new(max)

    fac[0]  = fac[1] = 1
    finv[0] = finv[1] = 1
    inv[1]  = 1
    2.upto(max-1) do |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
    end
  end

  def compute(n, k)
    return 0 if n < k
    return 0 if n < 0 || k < 0

    fac[n] * (finv[k] * finv[n - k] % mod) % mod
  end
end

N, M, K = gets.split.map(&:to_i)

MOD = 998244353
bc = BinomialCoefficients.new(2 * (10 ** 5), MOD)

res = 0
(K+1).times do |k|
  res += M * (M-1).pow(N-1-k, MOD) * bc.compute(N-1, k)
  res %= MOD
end

puts res