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)

# 繰り返し二乗法
def pow(base,exponent)
  raise ArgumentError if exponent < 0
  # 指数が0の時は結果は1です
  return 1 if exponent==0
  # 結果が出る変数を用意
  result = 1
  # 指数が1なら終了
  while exponent != 1 do
    # 指数が奇数の時、その時の底はここでresultに掛けてしまい、指数を一つ減らす
    if exponent.odd? then
      result *= base
      exponent -= 1
    end
    # 指数を2で割る
    exponent /= 2
    # 底の2乗を新たな底とする
    base *= base
    base = base % MOD
    # 繰り返す
  end
  result *= base
  # 結果を返す
  return result
end
MOD = 998244353

# これはべき場の高速化してない・・
# def inv(x)
#   res = 1 # 逆元をどんどんかけるやつ
#   beki = x # 逆元を計算したい値のMOD化でのべき乗
#   (MOD - 2).to_s(2).reverse.chars do |digest|
#     res = (beki * res) % MOD if digest == '1'
#     beki = (beki * beki) % MOD
#   end
#   res
# end



n,m,k = gets.chomp.split.map(&:to_i)

ncr = 1


# K組までのm*(m-1)を計算 1~n-1を計算しておいて、最後からk+1個を使えば良い
memo = [m%MOD]
1.upto(n-1) do |i|
  memo[i] = memo[i-1] * (m-1) % MOD
end

ans = memo.last

# n-1!まで計算
fac = Array.new(n, 1)
1.upto(n-1) do |i|
  fac[i] = fac[i-1] * i % MOD
end
# p fac

# rfac 逆元を計算したやつ
rfac = Array.new(n)
# 最も大きいのは最後のやつ
# mod pにおけるaの逆元はa^(p-2)つまり mod pの状況で bで割ってから pのあまりというのはb^(p-2)をかけるのに等しい
rfac[-1] = pow(fac[-1], MOD - 2)

# 全部で割ってしまってるので、かけていけば復元できる？
(n-2).downto(0) do |i|
  rfac[i] = rfac[i+1] * (i+1) % MOD
end

# p rfac

1.upto(k) do |i|
  ans += memo[n-i-1] * fac[n-1] * rfac[i] * rfac[n-i-1] % MOD
end



# n-1Ciを高速で計算して足していく
#
#
# ↓はあとでinvを高速化して試す
# 1.upto(k) do |i|
#   exp = memo[n-i-1]
#   if k == 1
#     ncr = (n-1)
#   else
#     ncr = ncr * (n-i) * inv(i)
#   end
#
#   ans += exp * ncr
# end
#
p ans % MOD