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)

#!/usr/bin/env python3
# vim: set fileencoding=utf-8

# pylint: disable=unused-import, invalid-name, missing-docstring, bad-continuation


"""Module docstring
"""

import functools
import heapq
import itertools
import logging
import math
import random
import string
import sys
from argparse import ArgumentParser
from collections import defaultdict, deque
from copy import deepcopy
from typing import Dict, List, Optional, Set, Tuple

MODULO = 998244353


def combi(n: int, k: int, fact: List[int]) -> int:
    r = 1
    for i in range(k):
        r *= n - i
    return (r // fact[k]) % MODULO


def factorial(n: int) -> List[int]:
    fact = [0] * (n + 1)
    inv = [0] * (n + 1)
    fact[0] = 1
    inv[0] = 1
    for i in range(1, n + 1):
        fact[i] = (i * fact[i - 1]) % MODULO
        inv[i] = pow(fact[i], MODULO - 2, MODULO)
    return fact, inv


def solve(n: int, m: int, k: int) -> int:
    result = 0
    # fill the cache
    fact, inv = factorial(n - 1)
    for i in range(k + 1):
        result = (
            result
            + (
                m
                * pow(m - 1, n - i - 1, MODULO)
                * fact[n - 1]
                * inv[i]
                * inv[n - 1 - i]
            )
        ) % MODULO
    return result


def do_job():
    "Do the work"
    LOG.debug("Start working")
    # first line is number of test cases
    N, M, K = map(int, input().split())
    result = solve(N, M, K)
    print(result)


def print_output(testcase: int, result) -> None:
    "Formats and print result"
    if result is None:
        result = "IMPOSSIBLE"
    print("Case #{}: {}".format(testcase + 1, result))
    # 6 digits float precision {:.6f} (6 is the default value)
    # print("Case #{}: {:f}".format(testcase + 1, result))


def configure_log() -> None:
    "Configure the log output"
    log_formatter = logging.Formatter("%(lineno)d - " "%(message)s")
    handler = logging.StreamHandler(sys.stdout)
    handler.setFormatter(log_formatter)
    LOG.addHandler(handler)


LOG = None
# for interactive call: do not add multiple times the handler
if not LOG:
    LOG = logging.getLogger("template")
    configure_log()


def main(argv=None):
    "Program wrapper."
    if argv is None:
        argv = sys.argv[1:]
    parser = ArgumentParser()
    parser.add_argument(
        "-v",
        "--verbose",
        dest="verbose",
        action="store_true",
        default=False,
        help="run as verbose mode",
    )
    args = parser.parse_args(argv)
    if args.verbose:
        LOG.setLevel(logging.DEBUG)
    do_job()
    return 0


if __name__ == "__main__":
    import doctest

    doctest.testmod()
    sys.exit(main())


class memoized:
    """Decorator that caches a function's return value each time it is called.
    If called later with the same arguments, the cached value is returned, and
    not re-evaluated.
    """

    def __init__(self, func):
        self.func = func
        self.cache = {}

    def __call__(self, *args):
        try:
            return self.cache[args]
        except KeyError:
            value = self.func(*args)
            self.cache[args] = value
            return value
        except TypeError:
            # uncachable -- for instance, passing a list as an argument.
            # Better to not cache than to blow up entirely.
            return self.func(*args)

    def __repr__(self):
        """Return the function's docstring."""
        return self.func.__doc__

    def __get__(self, obj, objtype):
        """Support instance methods."""
        return functools.partial(self.__call__, obj)