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)

#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>
#define FASTIO ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define rep(i, l, r) for (ll i = (l); i < (r); ++i)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pli pair<ll, int> 
#define pdd pair<ld, ld> 
#define pb push_back
#define reset(x,v) memset(x,v,sizeof(x))
#define mod 998244353
using namespace std;

ll power(ll a, ll b){
  if(b == 0 || a == 1)
    return 1;
  ll so = power(a, b /2);
  so =(so * so) % mod;
  if(b % 2 == 1)
    so = (so * a) % mod;
  return so;
}
ll fact[200001];
ll ncr(ll n, ll r){
  if(r > n)
    return 0;
  ll ans = fact[n];
  ans = (ans * power(fact[r], mod - 2)) % mod;
  ans = (ans * power(fact[n-r], mod - 2)) % mod;
  return ans;
}
int main() {
	FASTIO
	ll n, m, k;
	cin >> n >> m >> k;
  	fact[0] = 1;
  	rep(i, 1, 200001){
      fact[i] = (i * fact[i-1]) % mod;
    }
	ll ans = 0;
  	rep(i, 0, k + 1){
     ans = (ans + (((ncr(n - 1, i) * m) % mod) * power(m - 1, n - 1 - i)) % mod) % mod;
    }
  	cout << ans;
	return 0;
}