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: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#define MOD 998244353
#define int long long
using namespace std;
long long T, N, a, b, c, d, e, f;
int psum[200100];
int m(int a, int b = MOD - 2)
{
	if (b == 0)
		return 1;
	if (b % 2)
	{
		return a * m(a, b - 1) % MOD;
	}
	else
	{
		int aa = m(a, b / 2);
		return aa * aa % MOD;
	}
}
signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin >> a >> b >> c;
	int cc = 1;
	int ans = 0;
	int i;
	for (i = 0; i <= c; i++)
	{
		int p = b * m(b - 1, a - 1 - i)%MOD;
		p = p * cc % MOD;
		ans += p;
		cc *= (a - 1 - i);
		cc %= MOD;
		cc = cc * m(i + 1)%MOD;
	}
	cout << ans % MOD;
}