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;
#define rep(i,n) for (int (i) = 0; (i) <(n); (i)++) 
#define so(v) sort(v.begin(),v.end())
using ll = long long;
using vi=vector<int>;
using vii=vector<vi>;


const int MAX = 510000;
const int MOD = 998244353;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
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;
    }
}

// 二項係数計算
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;
}
ll f(ll n,ll m){
  //k=0の時の答え
  ll ans=m;
  rep(i,n-1){
    ans*=m-1;
    ans%=MOD;
  }
  return ans;
}
int main() {
    // 前処理
    COMinit();
    ll n,m,k;cin>>n>>m>>k;
    ll answer=0;
    ll ff=f(n-k,m);
    for(int i=k;i>=0;i--){
      answer+=(COM(n-1,i)%MOD)*(ff%MOD)%MOD;
      ff=(ff*(m-1))%MOD;
      answer%=MOD;
    }
    cout << answer << endl;
    //cout<<COM(n-1,1)%MOD<<endl;
}