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)

# frozen_string_literal: true

n, k = gets.split.map(&:to_i)
a = gets.split.map(&:to_i)
MOD = 10**9 + 7
def prod(a)
  p = 1
  a.each do |e|
    p *= e
    p %= MOD
  end
  p
end
aa = a.sort_by(&:abs).reverse

b = aa.take(k)

max = 0
if b.any?(0)
  max = 0
elsif b.count(&:negative?).even? || n == k
  max = prod(b)
else
  c = aa[k..-1]
  if c.any?(&:positive?)
    m = b.filter(&:negative?).max
    b.delete(m)
    b.push c.max
    max = prod(b)
  elsif b.any?(&:positive?)
    m = b.filter(&:positive?).min
    b.delete(m)
    b.push c.min
    max = prod(b)
  else
    max = prod(a.sort.reverse.take(k))
  end
end
puts max