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 Ruby (2.7.1) (judge verdict: Accepted)

# require 'pp'
# # IS_DEBUG = true
# IS_DEBUG = false
# def dputs str
# 	if(IS_DEBUG)
# 		puts str
# 	end
# end
# def dpp str
# 	if(IS_DEBUG)
# 		pp str
# 	end
# end
def comb(n,a)
	return $fact[n] * $finv[a] % $mod * $finv[n-a] % $mod;
end
def pow(a,n)
	prd = 1;
	exp = a;
	while(n > 0)
		if((n&1)==1)
			prd *= exp;
			prd %= $mod;
		end
		exp *= exp;
		exp %= $mod;
		n >>= 1;
	end
	return prd;
end
# constant
NMAX = 200015;
$mod = 998244353;
# preprocess
$fact = Array.new(NMAX);
$finv = Array.new(NMAX);
$fact[0] = 1;
$finv[0] = 1;
for i in 1..(NMAX)
	$fact[i] = $fact[i-1] * i % $mod;
	$finv[i] = pow($fact[i], $mod-2);
end

N,M,K=gets.chomp.split(' ').map{|n| n.to_i}

sum = 0
for k in 0..(K)
	sum += M * pow(M-1, N-1-k) * comb(N-1, k)
	sum %= $mod
end

puts sum