// https://atcoder.jp/contests/abc167/tasks/abc167_e
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define REP(i, n) FOR(i, 0, n)
#define REPR(i, n) for (int i = n - 1; i >= 0; i--)
#define FOR(i, s, n) for (int i = (s), i##_len = (n); i < i##_len; ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
#define DIV(a, b) ((a - 1) / b + 1)
const int MOD = 998244353;

template <int MOD> struct ModInt {
    static const int Mod = MOD;
    unsigned x;
    ModInt() : x(0) {}
    ModInt(signed sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    ModInt(signed long long sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) {
        if ((x += that.x) >= MOD)
            x -= MOD;
        return *this;
    }
    ModInt &operator-=(ModInt that) {
        if ((x += MOD - that.x) >= MOD)
            x -= MOD;
        return *this;
    }
    ModInt &operator*=(ModInt that) {
        x = (unsigned long long)x * that.x % MOD;
        return *this;
    }
    ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const {
        long long a = x, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b;
            std::swap(a, b);
            u -= t * v;
            std::swap(u, v);
        }
        return ModInt(u);
    }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt operator-() const {
        ModInt t;
        t.x = x == 0 ? 0 : Mod - x;
        return t;
    }
};
template <int MOD> ostream &operator<<(ostream &st, const ModInt<MOD> a) {
    st << a.get();
    return st;
};
template <int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
    ModInt<MOD> r = 1;
    while (k) {
        if (k & 1)
            r *= a;
        a *= a;
        k >>= 1;
    }
    return r;
}
using mint = ModInt<MOD>;
vector<mint> F_{1, 1}, R_{1, 1}, I_{0, 1};

void check_fact(int n) {
    for (int i = I_.size(); i <= n; i++) {
        I_.push_back(I_[MOD % i] * (MOD - MOD / i));
        F_.push_back(F_[i - 1] * i);
        R_.push_back(R_[i - 1] * I_[i]);
    }
}

mint I(int n) {
    check_fact(abs(n));
    return n >= 0 ? I_[n] : -I_[n];
}
mint F(int n) {
    check_fact(n);
    return n < 0 ? 0 : F_[n];
}
mint R(int n) {
    check_fact(n);
    return n < 0 ? 0 : R_[n];
}
mint C(int n, int r) { return F(n) * R(n - r) * R(r); }
mint P(int n, int r) { return F(n) * R(n - r); }
mint H(int n, int r) { return n == 0 ? (r == 0) : C(n + r - 1, r); }

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m, k;
    cin >> n >> m >> k;
    mint ans = 0, a = m;
    // 1番目はM通り選択できるが、2番目以降は M-1 通り選択できる。(前と異なる色)
    // 長さはNなので、2番目以降の数は N-1 となる。
    // 前と同じ色にする数をKとすると、N-1通りから連続する色を最大K種類選択する。(n-1)_C_(i)
    // 連続する色の数Kを増やすほど、2番目以降の数が少なくなるとみなせるので、
    // 2番目以降の数は N-1-i となる。
    REP(i, k + 1) { ans += C(n - 1, i) * a * ((a - 1) ^ (n - i - 1)); }
    cout << ans << endl;

    return 0;
}


Measured memory_bytes: 5788