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 Julia (1.4.0) (judge verdict: Wrong Answer)

const double = Float64
const int = Int64
const vint = Vector{int}
const vdouble = Vector{double}

int(x) = parse(int,x)
parseint(x) = parse(int, x)
readint() = map(parseint, split(readline()))

const MOD = 998244353
const NMax = 2*10^5

function main()
  
  # ここ lobal 変数
  fac = ones(int,NMax)
  inv = ones(int,NMax)
  finv = ones(int,NMax)
  for i in 2:NMax
    fac[i] = (fac[i-1]*i)%MOD
    inv[i] = ((1+div(MOD,i))*inv[i-MOD%i])%MOD
    finv[i] = (finv[i-1]*inv[i])%MOD
  end
  N,M,K = readint()
  
  if N == 1
    println(M)
  elseif M == 1
    println(1)
  else
    MM = Vector{int}(undef,N)
     MM[1] = M-1
    for i in 2:N
      MM[i] = (MM[i-1]*(M-1))%MOD
    end

    mod = 0
    for i in K:-1:1
      if N-1-i == 0
        mod += M
        mod %= MOD
      else
        mod += ((((fac[N-1]*finv[i]%MOD * finv[N-1-i]%MOD) * M) %MOD ) *MM[N-1-i]) %MOD
        mod %= MOD
      end
    end

    mod += (M * MM[N-1])%MOD
    mod %= MOD

    println(mod )
  end
end

main()