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<stdio.h>
#include<math.h>
int main(void){
	int n,m,k;
	long long int a,b,sum=0;
	scanf("%d %d %d",&n,&m,&k);
	
	for(int i=0;i<k+1;i++){
		//printf("%d",i);
		//sum+=((((long long int)pow(m-1,n-1-i))%998244353)*(((long long int)pow(m,1+i))%998244353))%998244353;
		a=1;b=1;
		for(int j=0;j<n-1-i;j++){
			a*=(m-1);
			a%=998244353;
		}
		for(int j=0;j<1+i;j++){
			b*=m;
			b%=998244353;
		}
		sum+=(a*b)%998244353;
		sum%=998244353;
	}
	printf("%lld",sum);
}