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>
#define ll long long int
#define M 1000000007
#define mod 998244353
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define pi pair<ll,ll>
#define endl "\n"
using namespace std;
const ll N=500010;

ll power(ll x,ll n)
{
    if(n==0){return 1;}
    if(n==1){return x;}
    if(n%2==0){
        return power((x*x)%mod,n/2);
    }
    if(n%2==1){
        return (power((x*x)%mod,n/2)*x)%mod;
    }
}

ll fac[N];

ll ncr(ll n,ll r)
{
    ll num=fac[n];
    ll denum=(fac[r]*fac[n-r])%mod;
    ll ans=(num*power(denum,mod-2))%mod;
    return ans;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    fac[0]=1;
    for(ll i=1;i<N;++i){
        fac[i]=(i*fac[i-1])%mod;
    }

    ll n,m,k;
    cin>>n>>m>>k;
    ll ans=0;
    for(ll b=1;b<=n;++b)
    {
        if(n-b<=k){
            ll ch=(m%mod*power((m-1+mod)%mod,b-1))%mod;
            ch=(ch*ncr(n-1,b-1))%mod;
            ans=(ans+ch)%mod;
        }
    }
    cout<<ans;

    return (0);
}