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 REP(i, s) for (int i = 0; i < s; ++i)
#define ALL(v) (v).begin(), (v).end()
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define DEBUG
#define int long long
#define INF 1e18
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, vector<vector<T> > P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }
template<class T> ostream& operator << (ostream &s, set<T> P)
{ EACH(it, P) { s << "<" << *it << "> "; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P)
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }
template<class T>void show(vector<T>v){for (int i = 0; i < v.size(); i++){cerr<<v[i]<<" ";}cerr<<"\n";}
typedef long long ll;

ll MOD = 998244353;

const int MAX = 510000;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit()
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll add(ll x, ll y)
{
    x += y;
    if (x >= MOD)
        return x - MOD;
    return x;
}
ll sub(ll x, ll y)
{
    x -= y;
    if (x < 0)
        return x + MOD;
    return x;
}
ll mult(ll x, ll y)
{
    return (x * y) % MOD;
}
ll bin_pow(ll x, ll p)
{
    if (p == 0)
        return 1;
    if (p & 1)
        return mult(x, bin_pow(x, p - 1));
    return bin_pow(mult(x, x), p / 2);
}

signed main()
{
    int n, m, k;
    cin >> n >> m >> k;
    COMinit();
    int cnt = 0;
    for (int i = 0; i <=k ; i++)
    {
        int x = COM(n - 1, i) * m;
        x %= MOD;
        cnt += x*bin_pow(m - 1, n - 1 - i);
        cnt %= MOD;
    }
    cout << cnt << endl;

    return 0;
}