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)

module ModInt
  MOD = 998244353

  def to_mod(a); a % MOD; end
  def add(a, b); (a + b) % MOD; end
  def mul(a, b); (a * b) % MOD; end
  def inv(a); a.pow(MOD - 2, MOD); end
  def div(a, b); mul(a, inv(b)); end
end

class FactTable
  include ModInt

  def initialize(n)
    @fact = [1] * (n + 1)
    @finv = [1] * (n + 1)
    2.upto(n) { |i| @fact[i] = mul(@fact[i - 1], i) }
    @finv[n] = inv(@fact[n])
    n.downto(3) { |i| @finv[i - 1] = mul(@finv[i], i) }
  end

  def nCk(n, m)
    mul(mul(@fact[n], @finv[m]), @finv[n - m])
  end
end

n,m,k = gets.split.map(&:to_i)
ft = FactTable.new(n)
include ModInt

ans = 0
0.upto(k) do |k|
  b = (m-1).pow(n-1-k, MOD)
  b = mul(b, m)
  b = mul(b, ft.nCk(n-1,k))
  ans = add(ans, b)
end
puts ans