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 C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Numerics;
using System.Linq;
using System.Collections.Generic;
using System.Text;

namespace debug
{

    class ProgramD
    {
        static void Main(string[] args)
        {
            //入力
            string[] s = Console.ReadLine().Split(' ');
            long n = long.Parse(s[0]);
            long m = long.Parse(s[1]);
            long k = long.Parse(s[2]);
            long MOD = 998244353;

            //下準備
            nCk com = new nCk();
            com.fac = new long[n + 1];
            com.inv = new long[n + 1];
            com.finv = new long[n + 1];
            com.mod = MOD;
            com.max = n + 1;

            //メモ作成
            com.com_init();

            long ans = 0;

            for(long i = 0; i <= k;i++)
            {
                if (i == 0)
                {
                    ans += modpow(m - 1, n - 1, MOD);
                    ans %= MOD;
                    continue;
                }

                ans += (com.com(n - 1, i) * modpow(m - 1, n - i - 1, MOD)) % MOD;
                ans %= MOD;

            }

            ans *= m;
            ans %= MOD;

            Console.WriteLine(ans);
                
        }

        //繰り返し二乗法で求める
        static long modpow(long a, long n, long MOD)
        {
            if (n == 0)
                return 1;
            if (n % 2 == 0)
            {
                long temp = modpow(a, n / 2, MOD);
                return temp * temp % MOD;
            }

            return a * modpow(a, n - 1, MOD) % MOD;
        }
    }

    class nCk
    {
        public long mod;
        public long max;
        public long[] fac;
        public long[] inv;
        public long[] finv;


        //メモ作成(前準備)
        public void com_init()
        {
            fac[0] = 1;
            fac[1] = 1;
            finv[0] = 1;
            finv[1] = 1;
            inv[1] = 1;
            for (long i = 2; i < max; i++)
            {
                fac[i] = i * fac[i - 1] % mod;
                inv[i] = mod - inv[mod % i] * (mod / i) % mod;
                finv[i] = finv[i - 1] * inv[i] % mod;
            }
        }

        //二項計算
        public long com(long n, long k)
        {
            if (n < 0 || k < 0)
                return 0;
            if (n < k)
                return 0;
            return fac[n] * (finv[n - k] * finv[k] % mod) % mod;
        }

    }
}