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>

using namespace std;
using LL = long long;

constexpr int MAXN = 2e5 + 5;
constexpr LL Mod = 998244353;

LL n, m, k, ans, fac[MAXN];

inline void Init() {
    fac[0] = 1;
    for (LL i = 1; i <= MAXN; i++) fac[i] = fac[i - 1] * i % Mod;
}

inline LL qpow(LL a, LL b) {  // a^b
    LL res = 1;
    while (b) {
        if (b & 1) res = res * a % Mod;
        a = a * a % Mod;
        b >>= 1;
    }
    return res;
}

inline LL C(LL N, LL M) {
    if (M > N || M < 0) return 0;
    return fac[N] * qpow(fac[M], Mod - 2) % Mod * qpow(fac[N - M], Mod - 2) % Mod;
}
// At most K adjacent pairs can be same. So at least n-1-k pairs have to be
// different. Let we need to make p pairs different. So it can be done in
// C(n-1,p)*M*(M-1)^p. Do this for all p from n-1-k to n-1.

//n - k 个数 相邻的数两两不同
int main() {
    // freopen("test.out", "w", stdout);
    //ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m >> k;
    Init();
    for (LL i = n - 1 - k; i <= n - 1; i++)
        ans = (ans + C(n - 1, i) * m % Mod * qpow(m - 1, i) % Mod) % Mod;
    cout << ans << endl;
    return 0;
}