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 ll long long int
#define pb push_back
#define mp make_pair
#define ios ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
const ll mod=998244353;
ll fct[200005];
int n,m;
ll power(ll a,ll b)
{
	if(a==0) return 0;
	ll ans=1;
	while(b>0) 
	{
		if(b%2==1) ans=(ans*a)%mod;
		a=(a*a)%mod;
		b/=2;
	}
	return ans;
}
void fact(int n)
{
	fct[0]=1; int i;
	for(i=1;i<=n;i++)
	{
		fct[i]=(fct[i-1]*i)%mod;
	}
}
ll inverse(ll x)
{
	return power(x,mod-2);
}
int main()
{
	int k,i;
	ll x,ans=0;
	cin>>n>>m>>k;
//	cout<<(m*call(1,k))%mod;
	fact(n);
//	ans=(m*power(m-1,n-1))%mod;
	for(i=0;i<=k;i++)
	{
		if(i>n-1) break;
		x=fct[n-1];
//		cout<<x<<" ";
		x=(x*(inverse(fct[n-1-i])))%mod; 
		x=(x*(inverse(fct[i])))%mod;
		x=(x*m)%mod;
		x=(x*power(m-1,n-1-i))%mod;
	
		ans=(ans+x)%mod;
//		ans+=(fct[n-1])*(inverse(fct[n-i-1]))*(inverse(fct[i]))*m
	}
	cout<<ans;
	return 0;
}