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 Python (3.8.2) (judge verdict: Accepted)

n, m, k = map(int, input().split())
mod = 998244353

def powerDX(n, r, mod):
  if r == 0: return 1
  if r%2 == 0:
    return powerDX(n*n % mod, r//2, mod) % mod
  if r%2 == 1:
    return n * powerDX(n, r-1, mod) % mod

def cmb(n, r, mod):
    if ( r<0 or r>n ):
        return 0
    r = min(r, n-r)
    return g1[n] * g2[r] * g2[n-r] % mod
 
g1 = [1, 1]
g2 = [1, 1]
inverse = [0, 1]
 
for i in range(2, n + 1 ):
    g1.append( ( g1[-1] * i ) % mod )
    inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
    g2.append( (g2[-1] * inverse[-1]) % mod )

ans = 0
for i in range(0, k+1):
  ans += m*cmb(n-1, i, mod)*powerDX(m-1, n-i-1, mod)
  ans %= mod
print(ans)