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: Accepted)

def pow(x, n):
    if n == 0:
        return 1
    res = pow(x * x % MOD, n//2)
    if n % 2 == 1:
        res = res * x % MOD
    return res

def cmb(n, r, p):
    if (r < 0) or (n < r):
        return 0
    r = min(r, n - r)
    return fact[n] * factinv[r] * factinv[n-r] % p
 
MOD = 998244353
N, M, K = map(int, input().split())

fact = [1, 1]  # fact[n] = (n! mod p)
factinv = [1, 1]  # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1]  # factinv 計算用
 
for i in range(2, N + 2):
    fact.append((fact[-1] * i) % MOD)
    inv.append((-inv[MOD % i] * (MOD // i)) % MOD)
    factinv.append((factinv[-1] * inv[-1]) % MOD)

ans = 0
Mm1 = M-1
Nm1 = N-1
Mm1pow = pow(Mm1, (N-K-1))
for k in range(K, -1, -1):
    ans = (ans + (M * cmb(Nm1, k, MOD) * Mm1pow)) % MOD
    Mm1pow = (Mm1pow * Mm1) % MOD
print(ans)