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)

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

def bin(a, b, p):
    res = 1
    while b > 0:
        if b & 1 > 0:
            res = res * a % p
        a = a * a % p
        b >>= 1

    return res

MOD = 998244353
N = 200000 + 50
f = [0 for i in range(N)]
invf = [0 for i in range(N)]

f[0] = invf[0] = 1
for i in range(1, N):
    f[i] = f[i-1] * i % MOD

invf[n] = bin(f[n], MOD-2, MOD)
for ri in range(1, N-1):
    i = n - ri
    invf[i] = invf[i+1] * (i+1) % MOD

def binom(n, m):
    if n < m or n < 0 or m < 0:
        return 0
    return f[n] * invf[m] % MOD * invf[n-m] % MOD

def g(n, m):
    return m * bin(m-1, n-1, MOD)

ans = 0
for i in range(0, k+1):
    # print(n-i, i, binom(n-1, i), bin(n-i, m, MOD), n-i, m)
    ans = (ans + binom(n-1, i) * g(n-i, m)) % MOD

print(ans)