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: Time Limit Exceeded)

use std::io;
use std::str::FromStr;
static MOD: u64 = 998244353;

fn input_tuple() -> (u64, u64, u64) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1], vec[2])
}

fn binary_power(x: u64, n: u64) -> u64 {
    if n == 0 {
      return 1;
    }
    if n % 2 == 0 {
      return (binary_power(x, n / 2) % MOD).pow(2) % MOD;
    } else {
      return binary_power(x, n - 1) % MOD * x % MOD;
    }
}
  
fn inverse_mod(n: u64) -> u64 {
    binary_power(n, MOD-2)
}  

fn combination(n: u64, mut m: u64) -> u64 {
    if n < m {
        return 0;
    }
    
    if m > n - m {
        m = n - m
    }

    let mut ret = 1u64;
    for i in 0..m {
        ret %= MOD;
        ret *= n - i;        
    }
    ret %= MOD;

    let mut div = 1u64;
    for i in 1..(m+1) {
        div %= MOD;
        div *= i;
    }
    div %= MOD;

    // println!("{}, {}", ret, div);
    ret * inverse_mod(div) % MOD
}

fn main() {
    let (n, m, k) = input_tuple();
    let mut ans = 0;
    for i in 0..(k+1) {
        let num = combination(n - 1, i) % MOD * m % MOD * binary_power(m - 1, n - 1 - i) % MOD; 
        // println!("{}", num);
        ans += num;
        ans %= MOD;
    }
    println!("{}", ans);
}