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: Runtime Error)

class Facts

  def initialize(mod=10**9+7, n_max=1)
    @mod = mod
    @n_max = n_max
    @fact = [1, 1]
    @inv = [0,1]
    @factinv = [1, 1]
    setup_table(n_max) if 1 < @n_max
  end

  def cmb(n,r)
    return 0 if r < 0 || n < r
    setup_table(n) if @n_max < n
    @fact[n] * (@factinv[r] * @factinv[n-r] % @mod) % @mod
  end

  def factorial(n)
    setup_table(n) if @n_max < n
    @fact[n]
  end

  def prm(n,k)
    setup_table(n) if @n_max < n
    @fact[n] * @factinv[n-k] % @mod
  end

  private
    # O(t)
    def setup_table(t)
      (@n_max+1).upto(t) do |i|
        @fact.push( @fact[-1] * i % @mod )
        @inv.push( -@inv[@mod % i] * (@mod / i) % @mod )
        @factinv.push( @factinv[-1] * @inv[-1] % @mod )
      end
      @n_max = t
    end
end

mod = 998244353

n, m, k = gets.to_s.split.map{|e|e.to_i}

b = Array.new(k+2)
b[-1] = m * (m-1).pow(n-2-k, mod)

ans = 0
f = Facts.new(mod)

k.downto(0) do |i|
  t = b[i] = b[i+1] * (m-1) % mod
  ans += f.cmb(n-1, i) * t % mod
end

puts ans