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>
#define ll long long
#include<vector>
 
using namespace std;
 
const ll mod = 998244353;
const ll N = 200000 + 10;
 
ll inv[N];
ll f[N];
 
ll ksm(ll a, ll b)
{
	ll rul = 1;
	while(b)
	{
		if(b & 1) rul = rul * a % mod;
		b >>= 1;
		a = a * a % mod;
	}
	return rul;
}
 
ll c(ll a, ll b)
{
	return f[a] * inv[a-b] % mod * inv[b] % mod; 
}
 
int main()
{
	ll n,m,k;
	cin>>n>>m>>k;
	
	int maxn = max(n,m);
	f[0] = 1;
	for(ll i = 1; i<=maxn; i++) f[i] = f[i-1] * i % mod;//求i! 
	inv[maxn] = ksm(f[maxn], mod-2);
	for(ll i = maxn-1; i>=0; i--) inv[i] = inv[i+1] * (i+1) % mod;//求逆元 
	
	ll ans = 0;//C(n-1,n-1-i)*m*(m-1)^(n-1-i) //n-1-i为隔板数 
	ll tem = ksm(m-1,n-1-k); 
	for(int i=n-1-k; i<=n-1; i++)
	{
		ans = ans + c(n-1, i) % mod * m % mod * tem % mod;
		ans = ans % mod;
		tem = tem * (m-1) % mod;
	}
	cout<<ans;
	return 0;
}