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::*;
const MOD: usize = 998244353;

fn mod_pow(mut x: usize, mut e: usize, m: usize) -> usize {
    let mut res = 1;
    while e > 0 {
        if e & 1 == 1 {
            res = res * x % m;
        }
        x = x * x % m;
        e >>= 1;
    }
    res
}

mod enumeration {
    pub struct Enumeration {
        fact: Vec<usize>,
        finv: Vec<usize>,
        modulo: usize,
    }
    impl Enumeration {
        pub fn new(n: usize, modulo: usize) -> Enumeration {
            let mut fact: Vec<usize> = vec![1; n];
            let mut inv: Vec<usize> = vec![1; n];
            let mut finv: Vec<usize> = vec![1; n];
            for i in 2..n {
                fact[i] = fact[i - 1] * (i as usize) % modulo;
                inv[i] = modulo - (modulo / (i as usize)) * inv[(modulo as usize % i)] % modulo;
                finv[i] = inv[i] * finv[i - 1] % modulo;
            }
            Enumeration {
                fact: fact,
                finv: finv,
                modulo: modulo,
            }
        }
        pub fn c(&self, n: usize, k: usize) -> usize {
            if n < k {
                return 0;
            }
            self.fact[n] * self.finv[n - k] % self.modulo * self.finv[k] % self.modulo
        }
    }
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let m: usize = itr.next().unwrap().parse().unwrap();
    let k: usize = itr.next().unwrap().parse().unwrap();

    let en = enumeration::Enumeration::new(n + 1, MOD);
    let mut ans = 0;
    for i in 0..k + 1 {
        ans = (ans + en.c(n - 1, i) * m % MOD * mod_pow(m - 1, n - (i + 1), MOD) % MOD) % MOD;
    }
    println!("{}", ans);
}