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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    final long MOD = 998244353;
    long n = sc.nextInt();
    long m = sc.nextInt();
    long k = sc.nextInt();
    long res = 0;
    long split = 1;
    for (long i = 1; i < n-k; i++) {
      split *= (n-i)*reverseMod(i, MOD)%MOD;
    }
    for (long i = n-k; i <= n; i++) {
      split *= (n-i)*reverseMod(i, MOD)%MOD;
      long color = m*(long)Math.pow(m-1, i-1)%MOD;
      res += split*color%MOD;
      res %= MOD;
    }
    System.out.println(res);
    sc.close();
  }

  static long reverseMod(long n, long mod) { // nのmodにおける逆元、すなわち1/nを求めるメソッド
    return modpow(n, mod-2, mod);
  }

  static long modpow(long a, long b, long mod) { // a*b(mod)を求めるメソッド
    long res = 1;
    while (b > 0) {
      if ((b & 1) == 1) {
        res = res * a;
      }
      a = a*a;
      b >>= 1;
    }
    return res;
  }
}