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 proconio::{fastout, input};

fn pow(mut x: u64, mut y: u64, z: u64) -> u64 {
    let mut a = 1;
    while y > 0 {
        if y & 1 == 1 {
            a = a * x % z;
        }
        x = x * x % z;
        y = y >> 1;
    }
    a
}

fn inv(x: u64, m: u64) -> u64 {
    pow(x, m - 2, m)
}

// fn comb(n: u64, r: u64, m: u64, fact: &Vec<u64>) -> u64 {
//     if r < 0 || r > n {
//         0
//     } else {
//         fact[n] * inv(fact[r], m) % M * inv(fact[n - r], m) % M
//     }
// }

#[fastout]
fn main() {
    input! {
        n: u64, m: u64, k: u64,
    }
    const M: u64 = 998244353;
    let mut fact = vec![0; 200_100];
    fact[0] = 1;
    for i in 1..fact.len() {
        fact[i] = i as u64 * fact[i - 1] % M;
    }
    let mut ans = 0;
    for i in 0..=k {
        let cnt = m * pow(m - 1, n - 1 - i, M) % M;
        let nn = (n - i) as usize;
        let rr = i as usize;
        let nr = nn + rr - 1;
        let comb = fact[nr] * inv(fact[rr], M) % M * inv(fact[nr - rr], M) % M;
        ans = (ans + cnt * comb % M) % M;
    }
    println!("{}", ans);
}