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{

    Scanner sc = new Scanner(System.in);

    int MOD = 998244353;

    public static void main(String[] args){
        new Main().run();
    }

    void run() {

        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();

        long ans = 0;
        long col = m;

        Combi cb = new Combi(n-1, MOD);


        for (int x=n-1; 0<=x; x--) {
            long now = col % MOD;
//            System.out.println(now);
            now *= cb.nck(n-1, x);
            now %= MOD;
//            System.out.println(now);
            if (x <= k) {
                ans += now;
                ans %= MOD;
            }
            col *= (m - 1);
            col %= MOD;
        }

        System.out.println(ans);

//        System.out.println(new Combi(n, MOD).facts[n]);


    }

    public class Combi {
        int n;   //　要素数
        int MOD;
        long[] facts;

        public Combi(int n, int MOD){
            this.n = n;
            this.MOD = MOD;
            facts = new long[n+1];
            facts[0] = 1;
            for (int i=1; i<n+1; i++){
                facts[i] = facts[i-1]*i%MOD;
            }
        }

        public long modpow(long a, long b){  // 計算量はlog(MOD);
            if (b==0) return 1;
            else if (b==1) return a;
            long x = modpow(a, b/2);
            return b%2==0 ? x*x%MOD : x*(x*a%MOD)%MOD;
        }
        public long inv(long n){ // 逆元。x^(-1)≡x^(p-2) (MOD p)
            return modpow(n, MOD-2);
        }
        public long nck(int n, int k){
            return n<k ? 0: facts[n]*(inv(facts[n-k])*inv(facts[k])%MOD)%MOD;
        }
    }

}