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: Time Limit Exceeded)

import math
mod = 998244353

def cached_modC(n, mod):
    fac = mod_factorial(mod)
    nf = fac(n)
    def modC(r):
        if r == 0 or r == n:
            return 1
        if r == 1 or r == n-1:
            return n
        nrf = fac(n-r)
        nr_pow = mod_pow(nrf, mod)
        rf = fac(r)
        r_pow = mod_pow(rf, mod)
        result = nf
        result *= nr_pow(mod-2)
        result %= mod
        result *= r_pow(mod-2)
        result %= mod

        return result

    return modC

def mod_factorial(mod):
    cache = [1,1,2]
    def _factorial(n):
        nonlocal cache
        if n == 0:
            return 1
        length = len(cache)
        v = cache[-1]
        while n >= length:
            v *= length 
            v %= mod
            cache.append(v)
            length +=1
        return cache[n]
    return _factorial

def mod_pow(x, mod):
    xx = [1, x]
    def _pow(y, x=x):
        nonlocal xx
        if y == 0:
            return 1
        while len(xx) <= math.log(y, 2)+1:
            v = xx[-1] * xx[-1]
            v %= mod
            xx.append(v)
        i = 1
        result = 1
        while y > 0:
            if y % 2 == 1:
                result *= xx[i]
                result %= mod
            y = y >> 1
            i += 1
        return result
    return _pow

def test():
    N, M, K = 3,2,1
    answer = 6
    v = resolve(N, M, K)
    print(f"N={N} M={M} K={K} {v}={answer}")
    N, M, K = 100,100,0
    answer = 73074801
    v = resolve(N, M, K)
    print(f"N={N} M={M} K={K} {v}={answer}")
    N, M, K = 60522,114575,7559
    answer = 479519525
    v = resolve(N, M, K)
    print(f"N={N} M={M} K={K} {v}={answer}")



def main():
    #test()
    N, M, K = map(int, input().split())
    v = resolve(N, M, K)
    print(v)


def resolve(N, M, K):
    result = 0
    m1pow = mod_pow(M-1, mod)
    modC = cached_modC(N-1, mod)
    for k in range(K+1):
        b = m1pow(N-1-k)
        c = modC(k)
        v =  b * c
        v %= mod
        #if k < 10:
        #    print(f"b={b} c={c} k={k} {v}")
        result += v
        result %= mod
    
    result *= M
    result %= mod

    result = int(result)

    return result

        
if __name__ == '__main__':
    main()