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>
#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define REPD(i, n) for (ll i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = a; i <= (ll)(b); i++)
#define FORD(i, a, b) for (ll i = a; i >= (ll)(b); i--)
#define ALL(x) (x).begin(), (x).end()
#define SIZE(x) ((ll)(x).size())
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))
#define INF 2e9
#define PI 3.14159265358979323846264338327950L
using namespace std;
typedef long long ll;

class mint
{
  const ll mod = 998244353;
  ll x;

public:
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const
  {
    return mint(-x);
  }
  mint &operator+=(const mint &a)
  {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint &a)
  {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint &a)
  {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint &a) const
  {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint &a) const
  {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint &a) const
  {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const
  {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }
  mint inv() const
  {
    return pow(mod - 2);
  }
  mint &operator/=(const mint &a)
  {
    return (*this) *= a.inv();
  }
  mint operator/(const mint &a) const
  {
    mint res(*this);
    return res /= a;
  }
  friend ostream &operator<<(ostream &os, const mint &m)
  {
    os << m.x;
    return os;
  }
};

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

  mint x = mint(m) * mint(m - 1).pow(n - 1);
  mint ans = x;
  FOR(i, 1, k)
  {
    x *= ((mint(n) - i) / m - 1) / i;
    ans += x;
  }

  cout << ans << endl;

  return 0;
}