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>
typedef long long int ll;
using namespace std;
#define sz 300009
#define mod 998244353
ll m,a,b,c,d,n,arr[sz],i,k,j,x,y,l;
ll brr[sz],vis[sz];
long long binpow(long long a,long long b){
  if(b<=0)
    return 1;
  long long res=binpow(a,b/2)%mod;
if(b%2){
    return ((((res%mod)*(res%mod))%mod)*a)%mod;
}
else
    return (res*res)%mod;
}
int main()
{
scanf("%lld %lld %lld",&n,&m,&k);
b=1;
j=1;
for(i=n;i>=n-k;i--)
{
c=binpow(m-1,i-1)*m;
c%=mod;
c*=b;
c%=mod;
b*=(i-1);
b%=mod;
b*=binpow(j,mod-2);
b%=mod;
a+=c;
j++;
}
printf("%lld\n",a);
return 0;
}