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 <iostream>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <algorithm>

typedef long long ll;
const ll MOD = 998244353;
const long long INF = 1LL<<60;
const double PI=3.14159265358979323846;
const ll NMAX=200005;
using namespace std;

ll pow(ll a, ll b) {
  if (b == 0) {
    return 1;
  } else if (b % 2 == 0) {
    ll d = pow(a, b / 2);
    return (d * d) % MOD;
  } else {
    ll d = pow(a, b - 1);
    return (a * d) % MOD;
  }
}

int main(){
    ll n,m,k;
    cin >> n >> m >> k;
    ll ans = 0;


    ll fact[NMAX+1];
    fact[0] = 1;
    fact[1] = 1;
    for(ll i = 2; i <= NMAX; i++){
        fact[i] = fact[i-1]*i;
        fact[i] %= MOD;
    }
    ll invfact[NMAX+1];
    invfact[NMAX] = pow(fact[NMAX], MOD-2);
    invfact[0] = 1;
    for(ll i = NMAX-1; i > 0; i--){
        if(i ==0) exit(1);
        invfact[i] = invfact[i+1]*(i+1)%MOD;
    }


    for(ll i = 0; i <= k; i++){
        if(n-i < i) break;
        ll tmp = m;
        tmp = tmp * pow(m-1,n-1-i) % MOD;
        if(n-i >= i){
            ll choose = fact[n-1]*invfact[n-1-i]%MOD;
            choose = choose*invfact[i]%MOD;
            tmp = tmp*choose %MOD;
        }
        ans = (ans + tmp) %MOD;
    }

    cout << ans%MOD << endl;
    return 0;
}