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>
#define pb push_back

using namespace std;
typedef long long LL;
const LL MOD = 998244353;
LL fak[200001];

LL pangkat(LL a, LL b) {
	if (b == 0) return 1;
	LL tmp = pangkat(a, b / 2) % MOD;
	tmp = (tmp * tmp) % MOD;
	if (b % 2) tmp *= a, tmp %= MOD;
	return tmp;
}

LL C(int n, int r) {
	if (r == 0) return 1;
	if (n < r) return 0;
	LL tmp = fak[n];
	tmp *= pangkat(fak[r], MOD - 2);
	tmp %= MOD;
	tmp *= pangkat(fak[n - r], MOD - 2);
	tmp %= MOD;
	//printf("%lld\n", tmp);
	return tmp;
}

int main() {
	int N, M, K;
	scanf("%d%d%d", &N, &M, &K);
	fak[0] = 1;
	for (int i = 1; i <= N; i++) {
		fak[i] = fak[i - 1] * i;
		fak[i] %= MOD;
	}
	LL ans = 0;
	int x = N;
	for (int i = 0; i <= K; i++) {
		ans += (((M * pangkat((M - 1), N - 1)) % MOD) * C(x - 1, i)) % MOD;
		ans %= MOD;
		N--;
		//printf("%lld\n", ans);
	}
	printf("%lld\n", ans);
}