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: Runtime Error)

import java.util.*;

public class Main {
	Scanner sc = new Scanner(System.in);
	final int MOD = 998244353;
	final int MAX = Integer.MAX_VALUE;
	final long LMAX = Long.MAX_VALUE;
	int len = (int)1e6 + 1;
	
	
	long[] fac, finv, inv;
	void doIt() {
		int N = sc.nextInt();
		int M = sc.nextInt();
		int K = sc.nextInt();
		comInit(N);
		long sum = 0;
		for(int i = 0; i <= K; i++) {
			long ans = 1;
			ans = ans * (long)M % MOD;
			ans = ans * modCom(N - 1, i) % MOD;
			ans = ans * modPow(M - 1, N - 1 - i) % MOD;
			sum = (sum + ans) % MOD;
		}
		System.out.println(sum);
	}
	
	long modPow(long a, long n) {
		long ans = 1;
		while(n > 0) {
			if((n & 1) == 1)
				ans = ans * a % MOD;
			a = a * a % MOD;
			n >>= 1;
		}
		return ans;
	}
	
	void comInit(int len) {
		fac = new long[len];
		finv = new long[len];
		inv = new long[len];
		fac[0] = fac[1] = 1;
	    finv[0] = finv[1] = 1;
	    inv[1] = 1;
	    for (int i = 2; i < len; i++){
	        fac[i] = fac[i - 1] * i % MOD;
	        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
	        finv[i] = finv[i - 1] * inv[i] % MOD;
	    }
	}
	
	long modCom(int n, int k){
	    if (n < k) return 0;
	    if (n < 0 || k < 0) return 0;
	    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
	}
	
	public static void main(String[] args) {
		new Main().doIt();
	}

}