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)

/*
Read the question and constraints carefully .Take pen and paper and start solving
*/

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,n) for(i=a;i<n;++i)
#define irep(i,n,a) for(i=n;i>a;--i)
#define mod 1000000007
#define pb push_back
#define big 9223372036854775807
#define big1 LONG_MAX
#define big2 ll_MAX
#define big3 1000000000
#define sma1 LONG_MIN
#define sma2 ll_MIN
#define sma3 -1000000000
#define mp make_pair
#define dub double
#define ivec vector<ll>
#define lvec vector<long long>
#define cvec vector<char>
#define svec vector<string>
#define mt make_tuple
#define MOD 998244353
#define pi acos(-1.0)/*
#define ss second.second
#define ff first.first
#define f first
#define s second
#define sf second.first
#define fs first.second*/
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)

ll p,ans;

ll modinv(ll num,ll md)
{
    p = md-2;
    ans = 1;

    for(;p;num=(num*num)%md,p/=2)
    {
        if(p&1)
            ans = (ans*num)%md;
    }

    return ans;

}

int main()
{

    ll n,m,k,i,ans=0,v,d;

    cin>>n>>m>>k;

    ll fac[n+1],p[n+1];

    fac[0]=1;

    for(i=1;i<=n;++i)
            fac[i] = (fac[i-1]*i)%MOD;

    p[0]=m;

    for(i=1;i<=n;++i)
            p[i] = (p[i-1]*(m-1))%MOD;


    for(i=n-1-k;i<=n-1;++i)
    {
        v = (fac[n-1]*p[i])%MOD;
        d = (fac[n-1-i]*fac[i])%MOD;


        v = (v*modinv(d,MOD))%MOD;

        ans = (ans+v)%MOD;

    }


    cout<<ans;


    return 0;
}