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;
typedef long long int lli;
#define MAX 2000000
#define MOD 1000000007
#define endl "\n"
lli N,K;
lli pref[MAX];
string arr[MAX];
lli factorial[MAX];
lli factinv[MAX];
vector <lli> iarr;
lli possum = 0;
lli ms = 0;
#define m 998244353  
// C function for extended Euclidean Algorithm 
lli gcdExtended(lli a, lli b, lli *x, lli *y) 
{ 
    if (a == 0) 
    { 
        *x = 0, *y = 1; 
        return b; 
    } 
    lli x1, y1; // To store results of recursive call 
    lli gcd = gcdExtended(b%a, a, &x1, &y1); 
    *x = y1 - (b/a) * x1; 
    *y = x1; 
  
    return gcd; 
}
lli modInverse(lli a) 
{ 
    lli x, y,res; 
    lli g = gcdExtended(a, m, &x, &y); 
    if (g != 1) 
        cout << "Inverse doesn't exist"; 
    
    else
    { 
        // m is added to handle negative x 
        res = (x%m + m) % m; 
    } 
    return res;
}  
long long binpow(long long a, long long b) {
    a %= m;
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}
lli nci(lli n, lli i){
    return factorial[n]*factinv[i]%m*factinv[n-i]%m;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    factorial[0] = 1;
    for (int i = 1; i <= MAX/10; i++) {
        factorial[i] = factorial[i - 1] * i % m;
     }
    for (int i = 1; i <= MAX/10; i++) {
        factinv[i] = modInverse(factorial[i]);
    }
   // cout<<nci(5,0);
   lli ans = 0;
    lli N,M,K;
    cin>>N>>M>>K;
    
    for(lli i = 0; i<=K; i++){
        lli qty = 0;
        qty += M*binpow(M-1,N-i-1);
        qty %=m;
        if(i != 0)
            qty *= nci(N-1,i);
        qty %=m;
        ans += qty;
        ans %=m;
        //cout<<ans<<endl;
    }
    cout<<ans<<endl;
    

}