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
module ModInt
  refine Integer do
    def +(b); (super b) % MOD; end
    def -(b); (super b) % MOD; end
    def *(b); (super b) % MOD; end
    def pow(b); super(b, MOD); end
    def /(b); self * b.pow(MOD - 2); end
  end
end
using ModInt

class FactTable
  attr_reader :fa, :fi
  def initialize(n)
    @fa = [1] * (n + 1)
    @fi = [1] * (n + 1)
    2.upto(n) { |i| fa[i] = fa[i - 1] * i }
    fi[n] = 1 / fa[n]
    n.downto(3) { |i| fi[i - 1] = fi[i] * i }
  end
 
  def nCk(n, m)
    fa[n] * fi[m] * fi[n - m]
  end
end
 
n,m,k = gets.split.map(&:to_i)
ft = FactTable.new(n)
 
ans = 0
0.upto(k) do |k|
  ans += (m-1).pow(n-1-k) * m * ft.nCk(n-1,k)
end
puts ans