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)

# -*- coding: utf-8 -*-


class FactMod():
    '''
    modの値が素数の時のfactと組み合わせを求める
    フェルマーの小定理を用いているため、modが素数以外の時は使えない
    '''

    def __init__(self, n, mod):
        '''
        コンストラクタ
        f:nまでの i!の値を　配列に入れる
        inv: (i!)^-1　の値を配列に入れる
        '''
        self.mod = mod
        self.f = [1]*(n+1)
        for i in range(1, n+1):
            self.f[i] = self.f[i-1]*i % mod

        self.inv = [pow(self.f[-1], mod-2, mod)]
        for i in range(1, n+1)[::-1]:
            self.inv.append(self.inv[-1]*i % mod)
        self.inv.reverse()

    def fact(self, n):
        '''
        n!の値を返す
        '''
        return self.f[n]

    def comb(self, n, r):
        '''
        nCrの値を返す
        '''
        ret = self.f[n] * self.inv[n-r]*self.inv[r]
        ret %= self.mod
        return ret

    def perm(self, n, r):
        """
        nPrの値を返す
        """
        ret = self.f[n] * self.inv[n-r]
        ret %= self.mod
        return ret
N,M,K = map(int,input().split())
MOD=998244353

F=FactMod(N,MOD)
ans=0
for k in range(K+1):
    ans+= M *pow(M-1,N-k-1,MOD)*F.comb(N-1,N-1-k)
    ans%=MOD
print(ans)