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 <float.h>
#include <inttypes.h>
#include <limits.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif

#define P 998244353

int k[200000];
int rk[200000];

int getk(int a, int x){
	if(x==1) return a;
	int b = getk(a, x/2);
	b = ((long)b)*b%P;
	if((x&1)==0){
		return b;
	}else{
		return ((long)b)*a%P;
	}
}
 
int comb(int a, int b){
	if(a<b) return 0;
	return ((long)k[a])*rk[b]%P*rk[a-b]%P;
}

void init(int n){
	k[0]=1;
	for(int i=1; i<=n; i++){
		k[i]=k[i-1]*((long)i)%P;
	}	
	rk[n]=getk(k[n], P-2);
	for(int i=n-1; i>=0; i--){
		rk[i]=((long)rk[i+1])*(i+1)%P;
	}
}
uint32_t nextpint(void){ char c=getchar_unlocked(); while(c<'0'||'9'<c) c=getchar_unlocked(); uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return x; }

int main(void){
	uint32_t n=nextpint();
	uint32_t m=nextpint();
	uint32_t k=nextpint();
	init(n-1);
	uint64_t ans=0;
	uint64_t sh = m--;
	for(uint32_t i=n-k-1; i>0; i--) sh=sh*m%P;
	for(uint32_t i=k; i>0; i--){
		uint64_t now = sh * comb(n-1, i) % P;
		ans+=now;
		sh=sh*m%P;
	}
	ans+=sh;
	printf("%"PRIu64"\n", ans%P);
}