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input

4 2 3

output

2

code

//Siberian Squirrel#include
   
    #define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)#define ACM_LOCALusing namespace std;typedef long long ll;const int N = 2e5 + 10;const int MOD = 998244353;ll n, m, k;int f[N];ll quick_pow(ll ans, ll p, ll res = 1) {
       for(; p; p >>= 1, ans = ans * ans % MOD)        if(p & 1) res = res * ans % MOD;    return res % MOD;}ll inv(ll ans) {
       return quick_pow(ans, MOD - 2);}ll C(int n, int m) {
       if(n < m || n <= 0 || m < 0) return 0;    if(m == 0) return 1;    return 1ll * f[n] * inv(f[m]) % MOD * inv(f[n - m]) % MOD;}inline void solve(ll res = 0, ll ans = 1) {
       if(!k)        res = m;    else        res = C(n - 1, k) * m % MOD * quick_pow(m - 1, k) % MOD;    cout << res << endl;}int main() {
       IO;#ifdef ACM_LOCAL    freopen("input.txt", "r", stdin);	freopen("output.txt", "w", stdout);#endif	f[0] = f[1] = 1;	for(int i = 2; i < N; ++ i) f[i] = 1ll * f[i - 1] * i % MOD;    int o = 1;//    cin >> o;    while(o --) {
           cin >> n >> m >> k;        solve();    }    return 0;}
   

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