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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define pan(i,n) for(int i=0;i<n;i++)
#define pans(i,n,a) for(int i=a;i<n;i++)
#define ll long long int
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define INF INT_MAX
#define PI 3.141592653
#define inf 2000007
#define pll pair<ll,ll>
#define pii pair<int,int>
ll gcd(ll a, ll b) { return a % b == 0 ? b : gcd(b, a % b); }
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }

int 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; // y = y/2 
        x = (x*x) % p; 
    } 
    return res; 
}
int modInverse(ll n, ll p) 
{ 
    return power(n, p-2, p); 
}
int nCrModPFermat(ll n, ll r, ll p) 
{
    if (r==0) 
      return 1;
    ll fac[n+1]; 
    fac[0] = 1; 
    for (ll i=1 ; i<=n; i++) 
        fac[i] = fac[i-1]*i%p; 
    return (fac[n]* modInverse(fac[r], p) % p * 
            modInverse(fac[n-r], p) % p) % p; 
} 
int main(){
    ll mod=998244353,n,m,k,ans=0;
    cin>>n>>m>>k;
    pan(i,k+1){
        ans=(ans%mod+(((m)%mod)*((nCrModPFermat(n-1,i,mod))%mod)*((power(m-1,n-1-k,mod))%mod))%mod+mod)%mod;
    }
    cout<<ans;
}