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>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll mod = 998244353;
const int MAX = 510000;

ll fac[MAX], finv[MAX], inv[MAX];

ll Pow(ll x, ll n){
  if(n == 0) return 1;
  if(n % 2 == 0) return Pow(x*x % mod, n/2);
  else return x * Pow(x,n-1) % mod;
}

void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++){
    fac[i] = fac[i-1] * (ll)i % mod;
    inv[i] = mod - inv[mod%i] * (mod/(ll)i) % mod;
    finv[i] = finv[i-1] * inv[i] % mod;
  }
}

ll COM(int n, int k){
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n-k] % mod) % mod;
}

int main(){
  ll n, m, k;
  cin >> n >> m >> k;
  if(m == 1) {
    if(k == n-1) cout << 1 << endl;
    else cout << 0 << endl;
    return 0;
  }
  COMinit();
  ll ans = 0;
  rep(i,k+1) {
    ll c = (COM(n-1,i) * m) % mod;
    (c *= Pow(m-1,n-1-i)) %= mod;
    (ans += c) %= mod; 
  }
  cout << ans << endl;
  return 0;
}