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: Runtime Error)

#include <iostream>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <iomanip>
#include <cmath>
typedef long long ll;
using namespace std;

ll mod = 998244353, u = 1, d[100001];

ll mypow(ll x, ll y){
  if(y == 1) return x;
  ll z = mypow(x, y / 2) % mod;
  if(y % 2 == 1){
    return ((z * z) % mod) * x % mod;
  }
  else{
    return (z * z) % mod;
  }
}

void pc(ll l){
  for(ll i = l; i >= 1; i--){
    u = (u * i) % mod;
  }
  ll x = 1;
  for(ll i = 1; i <= l; i++){
    x = (x * i) % mod;
    d[i] = mypow(x, mod - 2);
  }
}

ll choose(ll l, ll r){
  return ((u * d[r]) % mod) * d[l - r] % mod;
}

void solve(){
  ll n, m, k, x, y, z; cin >> n >> m >> k;
  x = (mypow(m - 1, n - 1) * m) % mod;
  ll ans = x;
  z = mypow(m - 1, n - k - 2);
  pc(n - 1);
  for(int i = k; i >= 1; i--){
    y = choose(n - 1, i);
    z = (z * (m - 1)) % mod;
    ans = (ans + (m * ((y * z) % mod) % mod)) % mod;
    //cout << y << " " << z << " " << ans << endl;
  }
  cout << ans << endl;
  return;
}

int main(){
  solve();
  return 0;
}