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 (ll i=0;i<(n);i++)
typedef long long ll;


ll modpow(ll a, ll n, ll mod){
    ll res=1;
    while(n>0){
        if(n&1) res = res*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return res;
}

ll modinv(ll a, ll mod){
    return modpow(a, mod-2, mod);
}


int main() {
    ll N,M,K,ans;
    cin>>N>>M>>K;
    ll mod = 998244353;
    ll kakuk = M*modpow(M-1,N-1,mod)%mod;
    ans = kakuk;
    ll gyakuM = modinv(M-1,mod);
    rep(i,K){
        kakuk = (kakuk * (N-i-1))%mod;
        kakuk = (kakuk * modinv(i+1,mod))%mod;
        kakuk = (kakuk * gyakuM)%mod; 
        ans += kakuk;
        ans = ans%mod;
    }
    if(M==1&&N-1==K) cout << 1 << endl;
    else cout << ans << endl;
}