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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii =  pair<int, int>;
using pll =  pair<long long, long long>;
constexpr char ln =  '\n';
constexpr long long MOD = 998244353;
constexpr long long INF = 1000000009LL;
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define fo(i,j,n) for (int i=(j); i < (n); i++)
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; }

const int MAX = 2100000;
ll 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;
}

ll rui(ll a,ll b){
     ll ans=1;
     while(b>0){
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b/=2;
    }
     return ans;
}

int main() {   
    COMinit();
    ll n, m, k; cin >> n >> m >> k;
    ll ans = 0;
    for(int i=0; i<=k; i++){
        ll res = 1;
        res = res * m % mod;
        ll ruijo = rui(m-1, n-1-i); 
        res = res * ruijo % mod;
        ll tmp = COM(n-1, i); 
        res = res * tmp % mod;
        ans = (ans + res) % mod;
    }
    cout << ans << ln;
}