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)

import math


class Combination:
    def __init__(self, n_max, mod=998244353):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * \
                modinv[self.mod % i] % self.mod
        return modinv


def power_func(a, n, p=998244353):
    bi = str(format(n, "b"))
    res = 1
    for i in range(len(bi)):
        res = (res*res) % p
        if bi[i] == "1":
            res = (res*a) % p
    return res


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

ans = 0
if n-1 == k:
    ans += m

for i in range(1, n):
    if n-i-1 <= k:
        ans += comb(n-1, i)*m*power_func(m-1, i)
        ans %= mod
print(ans % mod)