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;
#define rep(i,n) for(int i=0;i<n;i++)
#define rep2(i,a,b) for(int i=(a);i<(b);++i)

template<class T> inline void chmin(T& a, T b) {if (a > b) a = b;}
template<class T> inline void chmax(T& a, T b) {if (a < b) a = b;}


// combination
const int MAX = 510000;
const int MOD = 998244353;
 
long long fac[MAX], finv[MAX], inv[MAX];
 
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] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
long long 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;
}

int64_t modpow(int64_t x, int64_t n) {
    x = x%MOD;
    if(n==0) return 1;  //再帰の終了条件

    else if(n%2==1) {
        return (x*modpow(x, n-1))%MOD;  //nが奇数ならnを1ずらす
    }
    else return modpow((x*x)%MOD, n/2)%MOD;  //nが偶数ならnが半分になる
}

int main() {
  ll N,M,K;
  ll const mod = 998244353;
  cin >> N >> M >> K;
  ll ans=1;
  // ######;
  COMinit();
  vector<ll> memo(K+1);
  rep(i,K+1) {
    if (i==0) memo[i] = M*modpow(M-1,N-1)%mod;
    else {
      
      memo[i] = M*modpow(M-1,N-i-1)%mod*COM(N-1,i)%mod;
      // cout << COM(N-1,i) << endl;
      memo[i]%=mod;
      memo[i] += memo[i-1];
    }
    memo[i]%=mod;
  }
  // rep(i,K+1) cout << memo[i] << ' ';
  // cout << endl;
  if (memo[K]<0) memo[K]+=mod;
  cout << memo[K] << endl;
}