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 int long long
#define ll long long
#define pb push_back
#define mp make_pair
#define f(i,a,n) for(int i=a ; i<(int)n ; i++)
#define rf(i,n,a) for(int i=n ; i>=a ; i--)
#define F first
#define S second
#define all(c) (c).begin(),(c).end()
#define sz(v) (int)(v).size()
#define fast ios:: sync_with_stdio(false),cin.tie(nullptr);

typedef long double ld;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef pair<pii,int> ppi;
typedef vector<pii> vpi;

const int inf = 1e9;
const int inf64 = 1e18;
const int MOD = 998244353;
const int N = 2e5 + 5;
int fact[N], invfact[N];

int pow(int a, int b, int m)
{
   int ans=1;
   while(b)
   {
      if(b&1)
         ans=(ans*a)%m;
      b/=2;
      a=(a*a)%m;
   }
   return ans;
}

int modinv(int k)
{
   return pow(k, MOD-2, MOD);
}

void precompute()
{
   fact[0]=fact[1]=1;
   for(int i=2;i<N;i++)
   {
      fact[i]=fact[i-1]*i;
      fact[i]%=MOD;
   }
   invfact[N-1]=modinv(fact[N-1]);
   for(int i=N-2;i>=0;i--)
   {
      invfact[i]=invfact[i+1]*(i+1);
      invfact[i]%=MOD;
   }
}

int nCr(int x, int y)
{
   if(y>x)
      return 0;
   int num=fact[x];
   num*=invfact[y];
   num%=MOD;
   num*=invfact[x-y];
   num%=MOD;
   return num;
}

void solve(){
   int n,m,k;
   cin >> n >> m >> k;
   int ans = 0;
   for(int i=0 ; i<=k ; i++) {
      int K = n - i;
      int first = m;
      int extra = pow(m-1 , K - 1, MOD);

      int temp = (first * extra) %MOD;
      int C = nCr(n - 1 , K - 1);
      
      temp = (temp * C) %MOD;
      ans = (ans + temp) % MOD;
   }
   cout << ans;
}

int32_t main(){
   fast;
   precompute();
   int t = 1;
//   cin >> t;
   while(t--)  solve();   
}