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>
using namespace std;
#define ll long long
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
ll n, m, k;
ll Pow(ll a, ll b) {
	ll res = 1ll;
	while (b) {
		if (b & 1)res = res * a%mod;
		a = a * a%mod;
		b >>= 1;
	}
	return res;
}
ll fac[N], inv[N];
void init() {
	fac[0] = 1;
	for (ll i = 1; i < N; i++) {
		fac[i] = fac[i - 1] * i %mod;
	}
	inv[N - 1] = Pow(fac[N - 1], mod - 2);
	for (ll i = N - 2; i >= 0; i--) {
		inv[i] = inv[i + 1] * (i + 1) % mod;
	}
}
ll ans;
ll C(ll n, ll k) {
	return fac[n] * inv[k] % mod*inv[n - k] % mod;
}
int main() {
	init();
	scanf("%lld%lld%lld", &n, &m, &k);
	for (ll i = 0; i <= k; i++) {
		ll p = n - i;
		ans = (ans + C(n - 1, p - 1)*Pow(m - 1, n - i - 1) % mod*m%mod) % mod;
	}
	printf("%lld\n", ans);
	return 0;
}