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)

/*** author: yuji9511 ***/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using lpair = pair<ll, ll>;
const ll MOD = 998244353;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i=(m);i<(n);i++)
#define rrep(i,m,n) for(ll i=(m);i>=(n);i--)
#define printa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
void print() {}
template <class H,class... T>
void print(H&& h, T&&... t){cout<<h<<" \n"[sizeof...(t)==0];print(forward<T>(t)...);}
struct Combination{
private:
    ll N;
    vector<ll> fac, facinv;

public:
    Combination(ll n){
        N = n;
        fac.push_back(1); fac.push_back(1);
        rep(i,2,N+1) fac.push_back(fac[i-1] * i % MOD);
        rep(i,0,N+1) facinv.push_back(power(fac[i], MOD-2));
    }
    ll power(ll x, ll n){
        if(n == 0) return 1LL;
        ll res = power(x * x % MOD, n/2);
        if(n % 2 == 1) res = res * x % MOD;
        return res;
    }
    ll nck(ll n, ll k){
        if(k == 0 || n == k) return 1LL;
        return fac[n] * facinv[k] % MOD * facinv[n-k] % MOD;
    }
    ll npk(ll n, ll k){
        if(k == 0 || n == k) return 1LL;
        return fac[n] * facinv[n-k] % MOD;
    }
    ll get(ll x){return fac[x];};
    ll getinv(ll x){return facinv[x];};
};

void solve(){
    ll N,M,K;
    cin >> N >> M >> K;
    Combination cb(200010);
    ll ans = 0;
    rep(i,0,K+1){
        ll n = N - i;
        ll val = cb.nck(N-1, i);
        ll v = M;
        v *= cb.power(M-1, n-1);
        v %= MOD;
        ll res = v * val % MOD;
        ans += res;
        ans %= MOD;
    }
    print(ans);

    

}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
}