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<bits/stdc++.h>
using namespace std;
#define M_PI  3.1415926535
#define ll long long
#define ld long double
#define all(a) a.begin(),a.end()
#define Summon_Tourist ios::sync_with_stdio(false);cin.tie(0);
ll gcd(ll a , ll b) { return b==0?a: gcd( b, a%b); }
ll lcm(ll a , ll b) { return a/gcd(a,b) * b ; }
ll inf = 998244353;
ll modexp( ll base ,ll power)
 {
     if( power == 0  ) return 1;
     if( power & 1) return base*modexp( base , power-1 )%inf;
     return modexp(base*base%inf,power/2);
 }
///////////////////////////////////////--CKMKB--//////////////////////////////////////////////////
///////////////////////////////////////--PKMKB--//////////////////////////////////////////////////
 int main()
{
  Summon_Tourist
  //freopen("input.txt" , "r" , stdin );
  ll t = 1;
  //cin>>t;
   while(t--)
  {
    ll n,m,k;
    cin>>n>>m>>k;
    vector<ll> bincof(k+1,0);
    bincof[0] = 1 ;
    for( ll i = 1 ; i<=k ; i++ ){
        bincof[i] = ((bincof[i-1]%inf*(n-1-i+1)%inf)%inf*modexp(i,inf-2))%inf;
    }
    //for( ll i = 0 ; i<=k ; i++ ) cout<<bincof[i]<<" ";
    ll sum = 0 ;
    for( ll i =  0 ; i<=k ; i++ )
    {
         sum +=  ( m%inf * bincof[i]%inf * modexp( m-1 , n-i-1) )%inf;

    }
    cout<<sum%inf;

  }
  return 0;

}