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)

N, M, K = gets.split.map(&:to_i)
MOD = 998244353

result = 0

@fac = Array.new N+100
@inv = Array.new N+100
@finv = Array.new N+100

@fac[0] = @fac[1] = 1
@finv[0] = @finv[1] = 1
@inv[1] = 1

2.upto(N) 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


def comb n, k
  raise if n < k 
  raise if n < 0 || k < 0
  return @fac[n] * (@finv[k] * @finv[n-k] % MOD ) % MOD
end

0.upto(K) do |k|
  result += M * comb(N-1, k) * (M-1).pow(N-1-k, MOD)
  result %= MOD
end

puts result