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>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace __gnu_pbds;
using namespace std;
 
typedef   long long int ll;
typedef unsigned long long int sll;
typedef  double ld;
#define A 1000000007ll
#define B 998244353ll
#define C 1000000000000000000ll
#define FAST ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define oset tree<pair<ll,ll>, null_type,less<pair<ll,ll>>, rb_tree_tag,tree_order_statistics_node_update> 
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define mxe(v) *max_element(v.begin(),v.end())
#define mne(v) *min_element(v.begin(),v.end())
#define bs binary_search
#define lb lower_bound
#define ub upper_bound
#define ve vector
#define br break
#define PI 3.14159265358979323

ll power(ll x, ll y, ll p) 
{ 
    ll res = 1;
    x = x % p; 
    while (y > 0) 
    { 
        if (y & 1) 
            res = (res*x) % p; 
        y = y>>1; 
        x = (x*x) % p; 
    } 
    return res; 
} 

ll inv(ll n, ll p) 
{ 
    return power(n, p-2, p); 
} 
ll fac[300000];
ll ncr(ll n, ll r, ll p) 
{ 
    if (r==0)
    	return 1; 
    return (fac[n]* inv(fac[r], p) % p * inv(fac[n-r], p) % p) % p; 
} 

int main() 
{
    FAST;
    fac[0] = 1; 
    for (int i=1 ; i<300000; i++) 
        fac[i] = (fac[i-1]*i)%B; 
    ll n,m,k;
    cin>>n>>m>>k;
    ll ans = 0;
    for(int i=1;i<=k+1;i++)
    {
    	ll x = power(m-1,n-i,B);
    	x = (x*m)%B;
    	x = (x*ncr(n-1,i-1,B))%B;
    	ans = (ans+x)%B;
    }
    cout<<ans;
    //cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
    return 0;
}