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 d(x) cerr << #x ":" << x << endl;
#define dd(x, y) cerr << "(" #x "," #y "):(" << x << "," << y << ")" << endl
#define rep(i, n) for (int i = (int)(0); i < (int)(n); i++)
#define repp(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define all(v) v.begin(), v.end()
#define dump(v)                  \
    cerr << #v ":[ ";            \
    for (auto macro_vi : v) {    \
        cerr << macro_vi << " "; \
    }                            \
    cerr << "]" << endl;
#define ddump(v)                           \
    cerr << #v ":" << endl;                \
    for (auto macro_row : v) {             \
        cerr << "[";                       \
        for (auto macro__vi : macro_row) { \
            cerr << macro__vi << " ";      \
        }                                  \
        cerr << "]" << endl;               \
    }
using lint       = long long;
const int INF    = 1e9;
const lint LINF  = 1e18;
const lint MOD   = 1e9 + 7;
const double EPS = 1e-10;
const int mod    = 998244353;
struct mint {
    long long x;
    mint(long long x = 0) : x((x % mod + mod) % mod) {}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }
    mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2); }
    mint& operator/=(const mint a) { return (*this) *= a.inv(); }
    mint operator/(const mint a) const {
        mint res(*this);
        return res /= a;
    }
};
struct combination {
    vector<mint> 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;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i)
            ifact[i - 1] = ifact[i] * i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
    mint operator()(int n) { return fact[n]; }
};

int main() {
    lint N, M, K;
    cin >> N >> M >> K;
    combination comb(10000000);
    mint ans = 0;
    for (int k = 0; k <= K; k++) {
        int l = (N - 1) - k;
        // dd(k, l);
        // d(comb(M, l + 1).x);
        ans += comb(N - 1, l) * comb(M, 1) * comb(M - 1, 1).pow(l);
        // cout << ans.x << endl;
    }
    cout << ans.x << endl;
    return 0;
}