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: Accepted)

import java.util.*;

public class Main {

    private static final int MOD = 998244353;
    private static long[] fact;

    public static void main(String[] args) {


        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        int K = sc.nextInt();

        fact = new long[N];
        fact[0] = 1;
        for (int i = 1; i < N; i++) {
            fact[i] = i*fact[i-1];
            fact[i] %= MOD;
        }

        long ans = 0;
        long col = M;
        for (int x = N-1; x >= 0; x--) {
            if (x <= K) {
                long now = (col*comb(N-1,x))%MOD;
                ans += now;
                ans %= MOD;
            }
            col *= M-1;
            col %= MOD;
        }
        System.out.println(ans);
    }

    private static long comb(int n, int k) {
        return (fact[n]*pow(fact[k],MOD-2)%MOD)*pow(fact[n-k],MOD-2)%MOD;
    }

    private static long pow(long a, int b) {
        if (b == 0) {
            return 1;
        }
        if (b%2 == 0) {
            long half = pow(a,b/2);
            return half*half%MOD;
        } else {
            return pow(a,b-1)*a%MOD;
        }
    }
}