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: Wrong Answer)

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

long component(int n, int k);
long pow_amari(int x, int y, long key);

int main(void) {
  long n, m, k, i;
  long add = 0;
  long key = 998244353;

  scanf("%ld %ld %ld", &n, &m, &k);

  i = 0;
  while (i <= k) {
    add += component(n-i, i) * m * pow_amari(m-1, n-i-1, key);
    if (add >= key) add = add % key;
    i++;
  }

  printf("%ld\n", add);

  return 0;
}

long component(int n, int k) {
  long result = 1;
  int i = 0, multi;

  multi = n;
  while(i < k) {
    result *= multi;
    multi--;
    i++;
  }
  i = k;
  while(i > 0) {
    result /= i;
    i--;
  }

  return result;
}

long pow_amari(int x, int y, long key) {
  int i = 0;
  long result = 1;
  while (i < y) {
    result *= x;
    if (result >= key) {
      result = result % key;
    }
    i++;
  }
  return result;
}