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;
typedef long long ll;
const ll mod=998244353;
ll factorial[200001];
ll factorialinv[200001];
ll calc(ll a,ll b,ll c){
  if(b==0)return 1;
  else if(b%2==0){
    ll d=calc(a,b/2,c);
    return d*d%c;
  }
  else {
    return a*calc(a,b-1,c)%c;
  }
}
ll combi(ll a, ll b){
  return factorial[a]%mod*factorialinv[b]%mod*factorialinv[a-b]%mod;
}
int main(){
 long long int n,m,k,ans=0;
  cin >> n >>m >>k;
  factorial[0]=1;
  factorialinv[0]=1;
  for(ll i=1;i<=n;i++){
    factorial[i]=factorial[i-1]*i%mod;
    factorialinv[i]=factorialinv[i-1]*calc(i,mod-2,mod)%mod;
  }
  ll waru[n];
  ll temp;
  waru[0]=1;
  for(long long int i=1;i<n;i++)waru[i]=(waru[i-1]*(m-1))%mod;
  for(long long int i=n-1;i>=max(n-k-1,(long long)1);i--){
    temp=waru[i]*m%mod*combi(n-1,i); 
    ans=(ans+temp)%mod;
  }
  cout << ans <<endl;
}