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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

#左からn個目でendが特定の色の場合に
N,M,K = list(map(int,input().split()))

def xgcd(a, b):
  x0, y0, x1, y1 = 1, 0, 0, 1
  while b != 0:
    q, a, b = a // b, b, a % b
    x0, x1 = x1, x0 - q * x1
    y0, y1 = y1, y0 - q * y1
  return a, x0, y0
def modinv(a, m):
  g, x, y = xgcd(a, m)
  if g != 1:
    raise Exception('modular inverse does not exist')
  else:
    return x % m
def combination(n, k, mod=1e9+7) -> int:
  ret = 1
  for i in range(1, k+1):
    ret = (ret * (n-k+i) * modinv(i, mod))%mod
  return int(ret)

def pow(base, exponent, mod=1e9+7) -> int:
  if exponent == 0:
    return 1
  if exponent > 1:
    d,m = divmod(exponent, 2)
    return (base**m * pow(base, d, mod)**2 ) % mod
  else:
    return base

m = 998244353
res = 0
B = 1
if M != 1:
  mi = modinv(M-1,m)

  A = M * pow(M-1,N-1,m)
  for k in range(K+1):
    if k>0:
      B = (B*(N-k)*modinv(k,m))%m
    res = (res+ A*B)%m
    A = (A*mi)%m
  print (res)
else:
  print(1)