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)

from collections import deque
from heapq import heapify,heappop,heappush,heappushpop
from copy import copy,deepcopy
from itertools import product,permutations,combinations,combinations_with_replacement
from collections import defaultdict,Counter
from bisect import bisect_left,bisect_right
# from math import gcd,ceil,floor,factorial
# from fractions import gcd
from functools import reduce
from pprint import pprint

INF = float("inf")

def myinput():
    return map(int,input().split())

def mylistinput(n):
    return [ list(myinput()) for _ in range(n) ]

def mycol(data,col):
    return [ row[col] for row in data ]

def mysort(data,col,reverse_flag):
    data.sort(key=lambda x:x[col],reverse=reverse_flag)
    return data

def mymax(data):
    M = -1*float("inf")
    for i in range(len(data)):
        m = max(data[i])
        M = max(M,m)
    return M

def mymin(data):
    m = float("inf")
    for i in range(len(data)):
        M = min(data[i])
        m = min(m,M)
    return m

def mycount(ls,x):
    # lsはソート済みであること
    l = bisect_left(ls,x)
    r = bisect_right(ls,x)
    return (r-l)

def myoutput(ls,space=True):
    if space:
        if len(ls)==0:
            print(" ")
        elif type(ls[0])==str:
            print(" ".join(ls))
        elif type(ls[0])==int:
            print(" ".join(map(str,ls)))
        else:
            print("Output Error")
    else:
        if len(ls)==0:
            print("")
        elif type(ls[0])==str:
            print("".join(ls))
        elif type(ls[0])==int:
            print("".join(map(str,ls)))
        else:
            print("Output Error")

class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        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

n,m,k = myinput()
mod = 998244353

comb = Combination(n,mod)

ans = 0
for i in range(k+1):
    t = comb(n-1,i)
    ans1 = pow(m-1,n-1-i,mod)
    ans += ( m * t * ans1 ) % mod
print(ans%mod)