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 <bits/stdc++.h>
using namespace std;
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long

const int N = 2e5 + 5;
const int MOD = 998244353;

int pow(int a, int b, int m)
{
	int ans=1;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%m;
		b/=2;
		a=(a*a)%m;
	}
	return ans;
}

int n, m, k;
int cache[N], cache2[N];

int dp(int idx);

int dp2(int idx)
{
	if(idx > n + 1)
		return 0;
	int &ans = cache2[idx];
	if(ans != -1)
		return ans;
	ans = dp(idx);
	ans += dp2(idx + 1);
	ans %= MOD;
	return ans;
}

int dp(int idx)
{
	if(idx > n + 1)
		return 0;
	if(idx == n + 1)
		return 1;
	int &ans = cache[idx];
	if(ans != -1)
		return ans;
	ans = dp2(idx + 1) - dp2(idx + k + 2) + MOD;
	ans %= MOD;
	ans %= MOD;
	if(idx > 1)
		ans *= (m - 1);
	else
		ans *= m;
	ans %= MOD;
	return ans;
}

int32_t main()
{
	IOS;
	memset(cache, -1, sizeof(cache));
	memset(cache2, -1, sizeof(cache2));
	cin >> n >> m >> k;
	int ans = dp(1);
	ans %= MOD;
	cout << ans;
	return 0;
}