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 Kotlin (1.3.71) (judge verdict: Accepted)


import java.util.*

fun main() {
    val sc = Scanner(System.`in`)

    val n = sc.nextInt()
    val m = sc.nextInt()
    val k = sc.nextInt()
    val mod = 998244353L
    val facto = LongArray(n+1)
    val cache = HashMap<Pair<Int,Int>, Long>()
    val ex = LongArray(n+1){0L}

    ex[0] = 1L

    for(i in 1..n) {
        ex[i] = ex[i-1] * (m-1).toLong()
        ex[i] %= mod
    }

    facto[0] = 1L
    facto[1] = 1L

    for(i in 2..n) {
        facto[i] = (facto[i-1] * i) % mod
    }

    fun modInv(a:Long, m:Long = mod): Long {
        fun mInv(a:Long, m:Long): Long {
            if(m == 0L) return 1L
            if(m == 1L) return a
            var result = mInv(a, m / 2) % mod
            result = (result * result) % mod
            if(m % 2 != 0L) {
                result = (result * a) % mod
            }
            return result
        }
        return mInv(a, m - 2)
    }

    fun comb(n:Int, r:Int) : Long {
        val a = facto[n]
        val b = (facto[r] * facto[n-r]) % mod
        val ret = (a*modInv(b)) % mod
        return ret
    }

    var ans = 0L
    for(i in 0..k) {
        var v = comb(n - 1, i) * m.toLong()
        v %= mod
        v *= ex[n - 1 - i]
        v %= mod
        //println(v)
        ans += (v + mod) % mod
        ans %= mod
    }
    print(ans)


}