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;

typedef long long ll;
typedef pair<int, int> pii;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const int MOD = 998244353;
const int N = (int)2e5 + 10;

int powr(int a, int k){ 
  if(k == 0)
    return 1;
  int p = powr(a, k/2);
  p = (p * 1ll * p) % MOD;
  if(k % 2 == 1) p = (p * 1ll * a) % MOD;
  return p;
}

int fac[N];

int ncr(int n, int k){
  int r = fac[n];
  r=(r*1ll*powr(fac[k],MOD-2))%MOD;
  r=(r*1ll*powr(fac[n-k],MOD-2))%MOD;
  return r;
}

int main(){
  fastIO;
  fac[0]=1;
  for(int i = 1; i < N ; i ++ )
    fac[i]=(fac[i-1]*1ll*i)%MOD;
  int n, m, k;
  cin >> n >> m >> k;
  int ans = 0;
  int cur;
  for(int i = 0 ; i <= k ; i ++ ){
    cur = ncr(n - 1, i);
    cur = (cur * 1ll * m) % MOD;
    cur = (cur * 1ll * powr(m - 1, n - 1 - i)) % MOD;
    ans = (ans + cur) % MOD;
  }
  cout << ans << "\n";
  return 0;
}