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)

/*
SIGMA(C(n-1, k)*m*(m-1)^(n-k-1), k=0 to K)
*/

#include <bits/stdc++.h>

#define pb push_back

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

const ll MOD = 998244353;
const ll INF = 1<<30;

ll x, y;

ll ext_euclid(ll a, ll b){
	if(b==0){
		x = 1;
		y = 0;
		return a;
	}
	ll g = ext_euclid(b, a%b);
	/*
		bx + (a-a/b*b)y = 1	
		ay + b(x-a/b*y) = 1
	*/
	ll tmp = x;
	x = y;
	y = tmp-a/b*y;
	return g;
}

ll modinverse(ll a){
	ll g = ext_euclid(a, MOD);
	while(x<0)
		x+=MOD;
	return x;
}

ull powx(int b, ll e){
	if(e==0)
		return 1;
	ull ans = powx(b, e/2);
	ans = (ans*ans)%MOD;
	if(e%2)
		ans = (ans*b)%MOD;
	return ans;
}

int main(){
	ll n, m, k;
	vector<ull> fact;
	
	cin>>n>>m>>k;
	fact.assign(n+1, 1);

	for(int i=1;i<=n;i++)
		fact[i] = (fact[i-1]*i)%MOD;

	ull ans = 0;
	for(int i=0;i<=k;i++){
		ull tmp = m;
		tmp = (tmp*powx(m-1, n-i-1))%MOD;

		tmp = (tmp*fact[n-1])%MOD;
		tmp = (tmp*modinverse(fact[i]))%MOD;
		tmp = (tmp*modinverse(fact[n-1-i]))%MOD;

		ans = (ans+tmp)%MOD;
	}
	cout<<ans<<"\n";
	return 0;
}