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;
using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using pll = pair<ll, ll>;

const ll oo = 0x3f3f3f3f3f3f3f3fLL;
const double eps = 1e-9;

#define sz(c) ll((c).size())
#define all(c) begin(c), end(c)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define FORD(i, a, b) for (ll i = ll(b) - 1; i >= (a); i--)
#define TR(X)                                                  \
    ({                                                         \
        if (1) cerr << "TR: " << (#X) << " = " << (X) << endl; \
    })

constexpr ll MOD = 998244353ll;

ll exp(ll a, ll n) {
    ll res = 1;
    for (ll acc = a; n; acc = (acc*acc) % MOD, n>>=1) {
        if (n & 1) {
            res = (res * acc) % MOD;
        }
    }
    return res;
}

struct binom {
    vl fac, ifac;
    binom(ll N) : fac(N+1), ifac(N+1) {
        fac[0] = 1;
        FOR(i, 1, N+1) fac[i] = (fac[i-1] * i) % MOD;
        ifac[N] = exp(fac[N], MOD-2);
        FORD(i, 0, N) ifac[i] = (ifac[i+1] * (i+1)) % MOD;
    }

    ll get(ll n, ll k) {
        return (((fac[n] * ifac[k]) % MOD) * ifac[n-k]) % MOD;
    }
};

int main() { cin.sync_with_stdio(0);
    ll N, M, K; cin >> N >> M >> K;
    binom b(N);

    ll ans = 0;
    FOR(i, 0, K+1) {
        ans += ((b.get(N-1, i) * M) % MOD) * exp(M-1, N-1-i);
        ans %= MOD;
    }
    cout << ans << endl;
}  // cin.tie(0) bei schnellem Wechseln