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 Rust (1.42.0) (judge verdict: Accepted)

use std::io::{stdin, Read, StdinLock};
use std::str::FromStr;

#[allow(dead_code)]
struct Scanner<'a> {
    cin: StdinLock<'a>,
}
#[allow(dead_code)]
impl<'a> Scanner<'a> {
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner { cin: cin }
    }

    fn read<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        token.parse::<T>().ok()
    }

    fn input<T: FromStr>(&mut self) -> T {
        self.read().unwrap()
    }

    fn vec<T: FromStr>(&mut self, len: usize) -> Vec<T> {
        (0..len).map(|_| self.input()).collect()
    }

    fn mat<T: FromStr>(&mut self, row: usize, col: usize) -> Vec<Vec<T>> {
        (0..row).map(|_| self.vec(col)).collect()
    }
}

fn main() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin);

    let n: usize = sc.input();
    let m: usize = sc.input();
    let k: usize = sc.input();

    const MOD: usize = 998244353;
    let max_n: usize = 500_000;
    let com = Combination::new(max_n, MOD);

    let mut ans = 0usize;
    for i in 0..=k {
        ans += com.get(n - 1, i) * m % MOD
            * mod_pow(m - 1, n - i - 1, MOD) % MOD;
        ans %= MOD;
    }

    println!("{}", ans);
}
fn mod_pow(x: usize, n: usize, modulo: usize) -> usize {
    let mut n = n;
    let mut x = x;
    let mut ret = 1;
    while n > 0 {
        if n & 1 == 1 {
            ret = ret * x % modulo;
        }
        x = x * x % modulo;
        n >>= 1;
    }
    ret
}

struct Combination {
    fac: Vec<usize>,
    finv: Vec<usize>,
    modulo: usize,
}

impl Combination {
    fn new(max_n: usize, modulo: usize) -> Combination {
        let mut fac = vec![0; max_n];
        let mut finv = vec![0; max_n];
        let mut inv = vec![0; max_n];
        fac[0] = 1;
        fac[1] = 1;
        finv[0] = 1;
        finv[1] = 1;
        inv[1] = 1;
        for i in 2..max_n {
            fac[i] = fac[i - 1] * i % modulo;
            inv[i] = modulo - inv[modulo % i] * (modulo / i) % modulo;
            finv[i] = finv[i - 1] * inv[i] % modulo;
        }
        Combination {
            fac: fac,
            finv: finv,
            modulo: modulo,
        }
    }

    fn get(&self, n: usize, k: usize) -> usize {
        assert!(n >= k);
        self.fac[n] * (self.finv[k] * self.finv[n - k] % self.modulo) % self.modulo
    }

    #[allow(dead_code)]
    fn h(&self, n: usize, k: usize) -> usize {
        self.get(n + k - 1, k)
    }
}