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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int N = 2*1e5+5;
typedef long long ll;
vector<ll> fact(N,0);
ll mod= 998244353;
ll modpow(ll a,ll b){
        if(!a) return 0;
        ll res=1;a%=mod;
        while(b){
                if(b&1) { res=(res*a)%mod; }
                a=(a*a)%mod;
                b>>=1;
        }
        return res;
}
ll inv(ll a){
        return modpow(a,mod-2);
}
ll choose(ll a,ll b){
        if(a<0||b<0||a<b) return 0;
        ll val=(fact[a]*inv(fact[a-b]))%mod;
        return val;
}
void compute(){
        fact[0]=1;
        for(ll i=1;i<N;i++)
                fact[i]=(fact[i-1]*i)%mod;

}
int main(){
        ll n,m,k;
        cin>>n>>m>>k;
        compute();
        ll res=0;
        for(int i=0;i<=k;i++)
                res=(res+((m*choose(n-1,i))%mod*modpow(m-1,n-i-1))%mod)%mod;
        cout<<res<<endl;
}