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::input;

fn main() {
    input! {
        n: isize,
        m: isize,
        k: isize,
    }

    let div = 998244353;
    let max = 510000;

    let mut fac = vec![0; max];
    let mut finv = vec![0; max];
    let mut inv = vec![0; max];

    com_init(&mut fac, &mut finv, &mut inv, max, div);
    let mut ans = 0;

    let mut tmp = m;
    for _ in 0..n - k - 1 {
        tmp = tmp * (m - 1) % div;
    }

    for i in (0..=k).rev() {
        ans = (ans + com((n - 1) as usize, i as usize, &fac, &finv, div) * tmp % div) % div;
        tmp = tmp * (m - 1) % div;
    }
    println!("{}", ans);
}

fn com_init(
    fac: &mut Vec<isize>,
    finv: &mut Vec<isize>,
    inv: &mut Vec<isize>,
    max: usize,
    m: isize,
) {
    fac[0] = 1;
    fac[1] = 1;
    finv[0] = 1;
    finv[1] = 1;
    inv[0] = 0;
    inv[1] = 1;

    let mut i = 2;
    while i < max {
        fac[i] = fac[i - 1] * i as isize % m;
        inv[i] = m as isize - inv[m as usize % i] * (m / i as isize) % m;
        finv[i] = finv[i - 1] * inv[i] % m;
        i += 1;
    }
}

fn com(n: usize, k: usize, fac: &Vec<isize>, finv: &Vec<isize>, m: isize) -> isize {
    if n < k {
        return 0;
    }
    fac[n] * (finv[k] * finv[n - k] % m) % m
}