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 sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)


class Combination:
    def __init__(self, n_max, mod=10**9+7):
        # O(n_max + log(mod))
        self.mod = mod
        f = 1
        self.fac = fac = [f]
        for i in range(1, n_max+1):
            f = f * i % mod
            fac.append(f)
        f = pow(f, mod-2, mod)
        self.facinv = facinv = [f]
        for i in range(n_max, 0, -1):
            f = f * i % mod
            facinv.append(f)
        facinv.reverse()

    # "n 要素" は区別できる n 要素
    # "k グループ" はちょうど k グループ

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

    def C(self, n, r):
        if not 0 <= r <= n:
            return 0
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def P(self, n, r):
        if not 0 <= r <= n:
            return 0
        return self.fac[n] * self.facinv[n-r] % self.mod

    def H(self, n, r):
        if (n == 0 and r > 0) or r < 0:
            return 0
        return self.fac[n+r-1] * self.facinv[r] % self.mod * self.facinv[n-1] % self.mod

    def rising_factorial(self, n, r):  # 上昇階乗冪 n * (n+1) * ... * (n+r-1)
        return self.fac[n+r-1] * self.facinv[n-1] % self.mod

    def stirling_first(self, n, k):  # 第 1 種スターリング数  lru_cache を使うと O(nk)  # n 要素を k 個の巡回列に分割する場合の数
        if n == k:
            return 1
        if k == 0:
            return 0
        return (self.stirling_first(n-1, k-1) + (n-1)*self.stirling_first(n-1, k)) % self.mod

    def stirling_second(self, n, k):  # 第 2 種スターリング数 O(k + log(n))  # n 要素を区別のない k グループに分割する場合の数
        if n == k:
            return 1  # n==k==0 のときのため
        return self.facinv[k] * sum((-1)**(k-m) * self.C(k, m) * pow(m, n, self.mod) for m in range(1, k+1)) % self.mod

    def balls_and_boxes_3(self, n, k):  # n 要素を区別のある k グループに分割する場合の数  O(k + log(n))
        return sum((-1)**(k-m) * self.C(k, m) * pow(m, n, self.mod) for m in range(1, k+1)) % self.mod

    def bernoulli(self, n):  # ベルヌーイ数  lru_cache を使うと O(n**2 * log(mod))
        if n == 0:
            return 1
        if n % 2 and n >= 3:
            return 0  # 高速化
        return (- pow(n+1, self.mod-2, self.mod) * sum(self.C(n+1, k) * self.bernoulli(k) % self.mod for k in range(n))) % self.mod

    def faulhaber(self, k, n):  # べき乗和 0^k + 1^k + ... + (n-1)^k
        # bernoulli に lru_cache を使うと O(k**2 * log(mod))  bernoulli が計算済みなら O(k * log(mod))
        return pow(k+1, self.mod-2, self.mod) * sum(self.C(k+1, j) * self.bernoulli(j) % self.mod * pow(n, k-j+1, self.mod) % self.mod for j in range(k+1)) % self.mod

    def lah(self, n, k):  # n 要素を k 個の空でない順序付き集合に分割する場合の数  O(1)
        return self.C(n-1, k-1) * self.fac[n] % self.mod * self.facinv[k] % self.mod

    def bell(self, n, k):  # n 要素を k グループ以下に分割する場合の数  O(k**2 + k*log(mod))
        return sum(self.stirling_second(n, j) for j in range(1, k+1)) % self.mod


n, m, k = map(int, read().split())
mod = 998244353
comb = Combination(n, mod)
ans = 0
n -= 1
for i in range(k + 1):
    ans += comb.C(n, n - i) * m * pow(m - 1, n - i, mod) % mod
    ans %= mod
print(ans)