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>
#define SIZE 300010
using namespace std;
const long long MOD = 998244353;

long long fac[SIZE], inv[SIZE]; // fac 存储阶乘  inv 存储阶乘逆元

long long qpow(long long a, long long b, long long mod) { // 快速幂 a ^ b % mod
	long long ans = 1;
	while (b) {
		if (b & 1) ans = (ans * a) % mod;
		a = (a * a) % mod;
		b >>= 1;
	}
	return ans;
}

long long getInv(long long n, long long p) { // 求 n mod p 意义下的逆元
	return qpow(n, p - 2, p);
}

void getFacInv(int n, long long p) { // 阶乘逆元
	fac[1] = 1;
	for (long long i = 2; i <= n; ++i) fac[i] = fac[i - 1] * i % p;
	inv[n] = getInv(fac[n], p);
	for (long long i = n - 1; i >= 0; --i) inv[i] = inv[i + 1] * (i + 1) % p;
}

long long C(long long n, long long m) {
	return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr); cout.tie(nullptr);
	getFacInv(300000, MOD);
	long long N, M, K, ans = 0;
	cin >> N >> M >> K;
	
	for (int i = 0; i <= K; ++i) {
		ans += C(N - 1, N - i - 1) * qpow(M - 1, N - i - 1, MOD) % MOD;
		ans = (ans % MOD + MOD) % MOD;
	}

	cout << (ans * M % MOD + MOD) % MOD;
}