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 <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <string.h>
#include <vector>
#include <queue>
#include <cmath>
#include <bitset>
#include <complex>
#include <functional>
#include <numeric>
#include <iomanip>

#define SPBR(w, n) std::cout<<(w + 1 == n ? '\n' : ' ');
#define YES cout << "YES" << endl
#define Yes cout << "Yes" << endl
#define NO cout << "NO" << endl
#define No cout << "No" << endl
#define ALL(i) (i).begin(), (i).end()
#define FOR(i, a, n) for(int i=(a);i<(n);++i)
#define RFOR(i, a, n) for(int i=(n)-1;i>=(a);--i)
#define REP(i, n) for(int i=0;i<int(n);++i)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define IN(a, x, b) (a<=x && x<b)
#define OUT(a, x, b) (x<a || b<=x)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

#define int ll
using ll = long long;
using ull = unsigned long long;
using ld = long double;
/* const int MOD = 1000000007; */
const int MOD = 998244353;
const int INF = 1e18;
const double PI = acos(-1);

using namespace std;
struct INIT { INIT(){
  cin.tie(0); ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
}}INIT;


int powMod(int n, int p) {
  int res = 1;
  while (p) {
    if (p & 1) (res *= n) %= MOD;
    (n *= n) %= MOD;
    p >>= 1;
  }
  return res;
}
int invMod(int n) {
  return powMod(n, MOD - 2);
}

vector<int> fac, finv, inv;

void comb_init(int n) {
  fac  = vector<int>(n);
  finv = vector<int>(n);
  inv  = vector<int>(n);
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < n; 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;
  }
}
 
int comb(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;
}

signed main() {
  int N, M, K; 
  cin >> N >> M >> K;

  comb_init(400000);

  int ans = 0;
  REP(i, K+1){
    (ans += (M*powMod(M-1, N-1-i))%MOD*comb(N-1, i)) %= MOD;
  }

  cout << ans << "\n";

  return 0;
}