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)

MOD = 998244353

class ModCombination
  attr_reader :fac, :inv, :finv, :mod

  def initialize(n, mod: MOD)
    @mod = mod
    @fac = [1, 1]
    @inv = [1, 1]
    @finv = [1, 1]

    (2..n).each 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 combination(n, k)
    return 0 if n < k
    return 0 if n < 0 || k < 0

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

  def repeated_combination(n, k)
    combination(n + k - 1, k)
  end
end

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

total = 1
m_mod = [1]

N.times do
  total *= M
  total %= MOD

  m_mod << (m_mod.last * (M - 1)) % MOD
end

(K + 1).upto(N - 1) do |c|
  v = mc.combination(N - 1, c)

  total -= M * v * m_mod[N - 1 - c]
  total %= MOD
end

puts total