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 (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>

long modpow(long a, long n, long mod){
  long size;
  for (long i = 1; ; i++) {
    if (pow(2,i) > n) {
      size = i;
      break;
    }
  }
  long bit[size];
  bit[0] = n;
  for (long i = 1; i < size; i++) {
    bit[i] = bit[i-1]/2;
    bit[i-1] %= 2;
  }
  long powa[size];
  powa[0] = a;
  for (long i = 1; i < size; i++) {
    powa[i] = powa[i-1]*powa[i-1];
    powa[i] %= mod;
  }
  long answer = 1;
  for (long i = 0; i < size; i++) {
    if (bit[i] == 1) {
      answer *= bit[i]*powa[i];
      answer %= mod;
    }
  }
  return answer;
}
long modCombination(long n, long r, long mod) {
  long answer;
  if (n-r < r) {
    r = n-r;
  }
  long nu = 1;
  long de = 1;
  for (long i = 1; i <= r; i++) {
    nu *= n+1-i;
    nu %= mod;
    de *= i;
    de %= mod;
  }
  answer = nu;
  answer *= modpow(de,mod-2,mod);
  answer %= mod;
  return answer;
}

int main(void){

  long n,m,k;
  scanf("%ld %ld %ld", &n, &m, &k);
  long mod = 998244353;
  long answer = modpow(m-1,n-1,mod);
  answer *= m;
  answer %= mod;
  long plus;
  for (long i = 1; i <= k; i++) {
    plus = modCombination(n-1,i,mod);
    plus *= modpow(m-1,n-1-i,mod);
    plus %= mod;
    plus *= m;
    plus %= mod;
    answer += plus;
    answer %= mod;
  }
  printf("%ld\n", answer);

  return 0;
}