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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"io"
	"os"
	"strconv"
)

//const mod = 998244353

func main() {
	sc := newScanner(os.Stdin)
	N, M, K := sc.i(), sc.i(), sc.i()
	fac, ifac := factorial()
	sum := 0
	for k := 0; k <= K; k++ {
		com := combination(N-1, k, fac, ifac) * M % mod
		com = com * Pow(M-1, N-1-k) % mod
		sum = (sum + com) % mod
	}
	fmt.Println(sum)
}

// I/O
type scanner struct {
	sc *bufio.Scanner
}

func newScanner(input io.Reader) *scanner {
	sc := bufio.NewScanner(input)
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, 1024), int(1e+9))
	return &scanner{sc}
}

func (s *scanner) s() string {
	s.sc.Scan()
	return s.sc.Text()
}

func (s *scanner) i() int {
	i, e := strconv.Atoi(s.s())
	if e != nil {
		panic(e)
	}
	return i
}

func (s *scanner) f() float64 {
	f, e := strconv.ParseFloat(s.s(), 64)
	if e != nil {
		panic(e)
	}
	return f
}

func (s *scanner) bs() []byte {
	return []byte(s.s())
}

func (s *scanner) is(n int) []int {
	res := make([]int, n)
	for i := 0; i < n; i++ {
		res[i] = s.i()
	}
	return res
}

func (s *scanner) fs(n int) []float64 {
	res := make([]float64, n)
	for i := 0; i < n; i++ {
		res[i] = s.f()
	}
	return res
}

//func (m *math2) Pow(x, y int) int {
//	fx := float64(x)
//	fy := float64(y)
//	z := math.Pow(fx, fy)
//	return int(z)
//}
const mod = 998244353
const facNum = 300001

//const factorial = new
func Pow(a, n int) int {
	ans := 1
	for n > 0 {
		if (n & 1) == 1 {
			ans = ans * a % mod
		}
		a = a * a % mod
		n = n >> 1
	}
	return ans
}

func Gcd(a, b int) int {
	if a < b {
		return Gcd(b, a)
	}
	for b != 0 {
		a, b = b, a%b
	}
	return a
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func combination(n, k int, fac, ifac []int) int {
	if n < k || n < 0 {
		return 0
	}
	if k == 0 {
		return 1
	}
	//while n != 0
	ans := ifac[k] * ifac[n-k] % mod
	return ans * fac[n] % mod
}

// 階乗: factorial
// コンビネーションを計算する際に前もって計算しておく
// fac[k] => k! (mod M)
// ifac[k] => k!^(M-2) (mod M)
// n: ex. 2 * 10^5 => 200001
func factorial() (fac []int, ifac []int) {
	fac = make([]int, facNum)
	fac[0] = 1
	ifac = make([]int, facNum)
	ifac[0] = 1
	for i := 1; i < facNum; i++ {
		fac[i] = fac[i-1] * i % mod
		ifac[i] = ifac[i-1] * Pow(i, mod-2) % mod
	}
	return
}