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 <iostream>
#include <algorithm>
#define flush fflush(stdout)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<ll, ll> Pl;
const int mod = (int)998244353, INF = (int)1e9;
const int di[4] = { 1,0,-1,0 }, dj[4] = { 0,1,0,-1 };


#define N 200003
ll fac[N], inv[N], finv[N];

void cominit(int n) {
    int i;
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (i = 2; i < n + 2; 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;
    }
    return;
}

ll com(int n, int r) {
    if (n < r) return 0;
    if (n < 0 || r < 0) return 0;
    return (fac[n] * finv[r]) % mod * finv[n - r] % mod;
}

int main(void) {
    int n, m, k, i;
    ll fr[200003], ans;

	scanf("%d%d%d", &n, &m, &k);

    cominit(n);

    fr[0] = 1;
    for (i = 0; i <= n; i++) {
        fr[i + 1] = fr[i] * (ll)(m - 1) % mod;
    }

    ans = 0;
    for (i = 0; i <= k; i++) {
        ans = (ans + com(n - 1, i) * m % mod * fr[n - i - 1] % mod) % mod;
    }

    printf("%lld\n", ans);

	return 0;
}