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>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using bint = boost::multiprecision::cpp_int;
using ll = long long;
using pint = pair<int, int>;
using pll = pair<ll, ll>;
#define rep(i,n) for (int i = 0; i < (n); i++)
#define repSE(i,s,n) for (int i = (s); i < (n); i++)
#define rrepSE(i,s,e) for (int i = (s); i > (e); i--)
#define ssort(v) sort(v.begin(), v.end())
#define gsort(v) sort(v.rbegin(), v.rend())
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> bool chmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool chmin(T &m, const T q) { if (q < m) {m = q; return true;} else return false; }
#define DEBUG_PRINT(var) std::cout << #var << " = " << var << "\n";

const ll MOD = 998244353;

ll modpow(ll a, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

ll modinv(ll i, ll mod){
    return modpow(i, mod-2, mod);
}

struct combination {
    vector<ll> fact, ifact;
    combination(int n): fact(n+1), ifact(n+1) {
        assert(n < MOD);
        fact[0] = 1;
        for(int i=1; i<=n; i++) {
            fact[i] = (fact[i-1] * i) % MOD;
        }
        for(int i=0; i<=n; i++) {
            ifact[i] = modinv(fact[i], MOD);
        }
    }
    ll operator()(int n, int k) {
        if (k<0 || k>n) return 0;
        return (((fact[n]*ifact[k]) % MOD) * ifact[n-k]) % MOD;
    }
};

int main()
{
    int n, m, k;
    cin >> n >> m >> k;
    ll ans = 0;
    combination c(n);
    rep(i, k+1) {
        ll tmp = (c(n-1,i) * modpow(m-1, n-i-1, MOD)) % MOD;
        tmp = (m * tmp) % MOD;
        ans = (ans + tmp) % MOD;
    }
    cout << ans << "\n";
    return 0;
}