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;

const long long MOD = 998244353;

long long modpow(long long a, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}

long long modinv(long long a, long long mod) {
  return modpow(a, mod - 2, mod);
}

long long modadd(long long a, long long b, long long mod) {
  return ((a % mod) + (b % mod)) % mod;
}

long long modsub(long long a, long long b, long long mod) {
  return ((a % mod) - (b % mod)) % mod;
}

long long modmul(long long a, long long b, long long mod) {
  return ((a % mod) * (b % mod)) % mod;
}

long long moddiv(long long a, long long b, long long mod) {
  return ((a % mod) * modinv(b, mod)) % mod;
}


const int MAX = 510000;
long long fac[MAX], finv[MAX], inv[MAX];

void COMinit(long long mod) {
    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] * i % mod;
        inv[i] = mod - inv[mod%i] * (mod / i) % mod;
        finv[i] = finv[i - 1] * inv[i] % mod;
    }
}

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

void Main() {
  int N, M, K;
  cin >> N >> M >> K;

  if (N == 1) {
    cout << M << endl;
    return;
  }

  COMinit(MOD);

  long long ans = 0;
  long long num = M;
  for (int i = 0; i < N - K - 2; i++) {
    num = modmul(num, M - 1, MOD);
  }
  for (int i = K; i >= 0; i--) {
    num = modmul(num, M - 1, MOD);
    long long tmp = modmul(num, COM(N - 1, i, MOD), MOD);
    ans = modadd(ans, tmp, MOD);
  }
  cout << ans << endl;
}

int main() {
  std::cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  std::cout << std::fixed << std::setprecision(15);
  Main();
  return 0;
}