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: Runtime Error)

#include<bits/stdc++.h>

using namespace std;

#define INF 1000000007
#define LINF (1LL << 62)
#define PI 3.14159265358979

typedef long long i64;
typedef pair<i64,i64> P;

inline i64 mod(i64 a, i64 m) { return (a % m + m) % m; }

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }



i64 n, m, k;
int MOD = 998244353;
i64 mod_pow(i64 x, i64 y, i64 mod){
	i64 res = 1;
	while(y > 0){
		if(y & 1) res = res * x % mod;
		x = x*x %mod;
		y >>= 1;
	}
	return res;
}
const int MAX = 101010;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
	inv[0] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long C(i64 n, i64 k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
	/*i64 res = 1;
	for(int i = 1; i <= k; i++){
		res *= n-i+1;
		res %= MOD;
		res *= inv[i];
		res %= MOD;
	}
	return res;
	*/
}
void solve(){
	cin >> n >> m >> k;
	
	i64 ans = 0;
	COMinit();
	for(int i = 0; i <= k; i++){
		ans += m*mod_pow(m-1,n-i-1,MOD)%MOD * C(n-1,i) %MOD;
		ans %= MOD;
	}
	cout << ans << endl;
}

int main(){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	int t = 1;
	//cin >> t;
	while(t--){
		solve();
	}
	
	return 0;
}