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;

typedef long long ll;
typedef pair<int, int> ii;

const int N = 2e5 + 5;
const int MOD = 998244353;
int n, m, K;
ll fac[N];
ll inv[N];

ll power(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}

void init() {
  fac[0] = 1;
  for (int i = 1; i < N; ++i) {
    fac[i] = fac[i - 1] * i % MOD;
  }
  inv[N - 1] = power(fac[N - 1], MOD - 2);
  for (int i = N - 2; i >= 0; --i) {
    inv[i] = inv[i + 1] * (i + 1) % MOD;
  }
}

ll C(ll a, ll b) {
  return fac[a] * inv[a - b] % MOD * inv[b] % MOD;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();

  cin >> n >> m >> K;

  ll res = 0;
  for (int k = 0; k <= K; ++k) {
    res = (res + m * C(n - 1, k) % MOD * power(m - 1, n - 1 - k) % MOD) % MOD;
  }

  cout << res << endl;
  
  return 0;
}