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 PyPy2 (7.3.0) (judge verdict: Runtime Error)

from sys import stdin, stdout, setrecursionlimit
from functools import lru_cache

rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()
rli = lambda: map(int, stdin.readline().split())

INF, NINF = float('inf'), float('-inf')

def binom_tables(n, q):
	fact = [1 for _ in range(n + 1)]
	inv = [1 for _ in range(n + 1)]
	for i in range(1, n+1):
		fact[i] = (fact[i-1] * i) % q
		inv[i] = pow(fact[i], q-2, q)
	return fact, inv

def binom(n, k, q, F, I):
	return (F[n]*((I[k]*I[n-k])%q))%q

def main():
	MOD = 998244353
	n, m, k = rli()
	F, I = binom_tables(n, MOD)
	ans = 0
	for i in range(n-k, n+1):
		ans += (binom(n-1, i-1, MOD, F, I)*m*pow(m-1, i-1, MOD))%MOD
	print(ans % MOD)

if __name__ == "__main__":
	main()